13. Implicit differentiation and the second order derivatives

수학 모음 (Maths collection) 전체보기

 

Question 1. Given $x^2+y^2=r^2$, show that $$ \begin{align} \frac{{\rm d}y}{{\rm d}x}&=-\frac{x}{y} \\ \\ \frac{{\rm d}^2y}{{\rm d}x^2}&=-\frac{r^2}{y^3} \end{align} $$

 

Solution.

We use implicit differentiation which is an application of chain rule and product rule as we will see here.

$$ \begin{align} && x^2+y^2&=r^2 \\ &\Rightarrow& \frac{{\rm d}}{{\rm d}x} \left( x^2+y^2 \right) &=\frac{{\rm d}}{{\rm d}x}r^2 \\ &\Rightarrow& \frac{{\rm d}}{{\rm d}x}x^2 + \frac{{\rm d}}{{\rm d}x}y^2 &=0 \\ &\Rightarrow& 2x+\underbrace{\frac{{\rm d}y}{{\rm d}x}\frac{{\rm d}}{{\rm d}y}}_{{\rm chain\;rule}}y^2&=0 \qquad{\rm by\;chain\;rule} \\ &\Rightarrow& 2x+2y\frac{{\rm d}y}{{\rm d}x}&=0 \\ &\Rightarrow& x+y\frac{{\rm d}y}{{\rm d}x}&=0 \\ &\Rightarrow& \frac{{\rm d}y}{{\rm d}x}&=-\frac{x}{y}=-\frac{x}{\pm\sqrt{r^2-x^2}}. \qquad\square \end{align} $$

 

For the second derivative, there are two different ways to find it.

 

(i) Method 1: By quotient rule. $$ \begin{align} && \frac{{\rm d}}{{\rm d}x}\left( \frac{u}{v} \right)&=\frac{u'v-uv'}{v^2} \\ \\ &\Rightarrow& \frac{{\rm d}^2y}{{\rm d}x^2}=\frac{{\rm d}}{{\rm d}x}\left( -\frac{x}{y} \right) &=-\frac{\left(\frac{{\rm d}}{{\rm d}x}x\right)y - x\left( \frac{{\rm d}}{{\rm d}x}y\right)}{y^2} \\ &&&=-\frac{y - x\frac{{\rm d}y}{{\rm d}x}}{y^2} \\ &&&=-\frac{y - x \left( -\frac{x}{y} \right)}{y^2} \\ &&&=-\frac{y + \frac{x^2}{y}}{y^2} \\ &&&=-\frac{x^2+y^2}{y^3} \\ &&&=-\frac{r^2}{y^3}. \qquad\square \end{align} $$

 

(ii) Method 2: By implicit differentiation. We start with: $$ \begin{align} && x+y\frac{{\rm d}y}{{\rm d}x}&=0 \\ &\Rightarrow& \frac{{\rm d}}{{\rm d}x} x+ \frac{{\rm d}}{{\rm d}x} \left( y\frac{{\rm d}y}{{\rm d}x} \right) &=0 \\ &\Rightarrow& 1 + \underbrace{\left( \frac{{\rm d}}{{\rm d}x}y \right)\frac{{\rm d}y}{{\rm d}x} + y\frac{{\rm d}^2y}{{\rm d}x^2}}_{\rm product\;rule}&=0 \qquad{\rm by\;product\;rule} \\ &\Rightarrow& 1 + \left( \frac{{\rm d}y}{{\rm d}x} \right)^2 + y\frac{{\rm d}^2y}{{\rm d}x^2}&=0 \\ &\Rightarrow& 1 + \left( \frac{{\rm d}y}{{\rm d}x} \right)^2 + y\frac{{\rm d}^2y}{{\rm d}x^2}&=0 \\ \\ &\Rightarrow& y\frac{{\rm d}^2y}{{\rm d}x^2}&=-\left[ 1 + \left( \frac{{\rm d}y}{{\rm d}x} \right)^2 \right] \\ &&&=-\left[ 1 + \left( -\frac{x}{y} \right)^2 \right] \\ &&&=-\left( 1 + \frac{x^2}{y^2} \right) \\ &&&=-\frac{x^2+y^2}{y^2} \\ &&&=-\frac{r^2}{y^2} \\ \\ && \Rightarrow\qquad \frac{{\rm d}^2y}{{\rm d}x^2} &=-\frac{r^2}{y^3}. \qquad\square \end{align} $$

 

Question 2. Given $x^2+4y^2=a^2$, show that $$ \begin{align} \frac{{\rm d}y}{{\rm d}x} & = -\frac{x}{4y} \\ \\ \frac{{\rm d}^2y}{{\rm d}x^2}&=-\frac{1}{16y^3} \end{align} $$

 

Solution.

We use implicit differentiation which is an application of chain rule and product rule as we will see here.

$$ \begin{align} && x^2 + 4y^2 &= a^2 \\ &\Rightarrow& \frac{{\rm d}}{{\rm d}x} \left( x^2 + 4y^2 \right) &=\frac{{\rm d}}{{\rm d}x} a^2 \\ &\Rightarrow& \frac{{\rm d}}{{\rm d}x}x^2 + \frac{{\rm d}}{{\rm d}x}y^2 &=0 \\ &\Rightarrow& 2x + \underbrace{ \frac{{\rm d}y}{{\rm d}x} \frac{{\rm d}}{{\rm d}y} }_{ {\rm chain\;rule} } 4y^2 &=0 \qquad{\rm by\;chain\;rule} \\ &\Rightarrow& 2x + 8y \frac{{\rm d}y}{{\rm d}x} &= 0 \\ &\Rightarrow& x + 4y \frac{{\rm d}y}{{\rm d}x} &= 0 \\ &\Rightarrow& \frac{{\rm d}y}{{\rm d}x} &= - \frac{ x }{ 4y } = - \frac{x}{ \pm 2 \sqrt{a^2-x^2} }. \qquad\square \end{align} $$

 

For the second derivative, there are two different ways to find it.

 

(i) Method 1: By quotient rule. $$ \begin{align} && \frac{{\rm d}}{{\rm d}x}\left( \frac{u}{v} \right) &= \frac{u'v-uv'}{v^2} \\ \\ &\Rightarrow& \frac{{\rm d}^2y}{{\rm d}x^2} = \frac{{\rm d}}{{\rm d}x}\left( -\frac{x}{4y} \right) &= - \frac{ \left(\frac{{\rm d}}{{\rm d}x}x\right)4y - x\left( \frac{{\rm d}}{{\rm d}x}4y\right) }{16y^2} \\ &&&= - \frac{ y - x \frac{{\rm d}y}{{\rm d}x} }{ 4y^2 } \\ &&&= - \frac{ y - x \left( - \frac{x}{4y} \right) }{ 4y^2 } \\ &&&= - \frac{ y + \frac{ x^2 }{ 4y } }{ 4y^2 } \\ &&&= - \frac{ x^2 + 4y^2 }{ 16y^3 } \\ &&&= - \frac{ a^2 }{ 16y^3 }. \qquad\square \end{align} $$

 

(ii) Method 2: By implicit differentiation. We start with: $$ \begin{align} && x + 4y \frac{{\rm d}y}{{\rm d}x} &= 0 \\ &\Rightarrow& \frac{{\rm d}}{{\rm d}x} x + \frac{{\rm d}}{{\rm d}x} \left( 4y \frac{{\rm d}y}{{\rm d}x} \right) &=0 \\ &\Rightarrow& 1 + \underbrace{ \left( \frac{{\rm d}}{{\rm d}x}4y \right)\frac{{\rm d}y}{{\rm d}x} + 4y\frac{{\rm d}^2y}{{\rm d}x^2} }_{\rm product\;rule}&=0 \qquad{\rm by\;product\;rule} \\ &\Rightarrow& 1 + 4 \left( \frac{{\rm d}y}{{\rm d}x} \right)^2 + 4y \frac{{\rm d}^2y}{{\rm d}x^2}&=0 \\ \\ &\Rightarrow& 4y \frac{{\rm d}^2y}{{\rm d}x^2} &= - \left[ 1 + 4 \left( \frac{{\rm d}y}{{\rm d}x} \right)^2 \right] \\ &&&= - \left[ 1 + 4 \left( -\frac{x}{4y} \right)^2 \right] \\ &&&= - \left( 1 + \frac{ x^2 }{ 4y^2 } \right) \\ &&&= - \frac{ x^2 + 4y^2 }{ 4y^2 } \\ &&&= - \frac{ a^2 }{ 4y^2 } \\ \\ && \Rightarrow\qquad \frac{{\rm d}^2y}{{\rm d}x^2} &= - \frac{ a^2 }{ 16y^3 }. \qquad\square \end{align} $$

 

Question 3. Given $3x^2+y^2=a^2$, show that $$ \begin{align} \frac{{\rm d}y}{{\rm d}x}&=-\frac{3x}{y} \\ \\ \frac{{\rm d}^2y}{{\rm d}x^2}&=-\frac{3a^2}{y^3} \end{align} $$

 

Solution.

We use implicit differentiation which is an application of chain rule and product rule as we will see here.

$$ \begin{align} && 3x^2+y^2&=a^2 \\ &\Rightarrow& \frac{{\rm d}}{{\rm d}x} \left( 3x^2+y^2 \right) &=\frac{{\rm d}}{{\rm d}x}a^2 \\ &\Rightarrow& \frac{{\rm d}}{{\rm d}x}\left(3x^2\right) + \frac{{\rm d}}{{\rm d}x}y^2 &=0 \\ &\Rightarrow& 6x+\underbrace{\frac{{\rm d}y}{{\rm d}x}\frac{{\rm d}}{{\rm d}y}}_{{\rm chain\;rule}}y^2&=0 \qquad{\rm by\;chain\;rule} \\ &\Rightarrow& 6x+2y\frac{{\rm d}y}{{\rm d}x}&=0 \\ &\Rightarrow& 3x+y\frac{{\rm d}y}{{\rm d}x}&=0 \\ &\Rightarrow& \frac{{\rm d}y}{{\rm d}x}&=-\frac{3x}{y}. \qquad\square \end{align} $$

 

For the second derivative, there are two different ways to find it.

 

(i) Method 1: By quotient rule. $$ \begin{align} && \frac{{\rm d}}{{\rm d}x}\left( \frac{u}{v} \right)&=\frac{u'v-uv'}{v^2} \\ \\ &\Rightarrow& \frac{{\rm d}^2y}{{\rm d}x^2}=\frac{{\rm d}}{{\rm d}x}\left( -\frac{3x}{y} \right) &=-\frac{\left(\frac{{\rm d}}{{\rm d}x}3x\right)y - 3x\left( \frac{{\rm d}}{{\rm d}x}y\right)}{y^2} \\ &&&=-\frac{3y - 3x\frac{{\rm d}y}{{\rm d}x}}{y^2} \\ &&&=-3\left[\frac{y - x \left( -\frac{3x}{y} \right)}{y^2}\right] \\ &&&=-3\left[\frac{y + \frac{3x^2}{y} }{y^2}\right] \\ &&&=-3\left[\frac{3x^2+y^2}{y^3}\right] \\ &&&=-\frac{3a^2}{y^3}. \qquad\square \end{align} $$

 

(ii) Method 2: By implicit differentiation. We start with: $$ \begin{align} && 3x+y\frac{{\rm d}y}{{\rm d}x}&=0 \\ &\Rightarrow& \frac{{\rm d}}{{\rm d}x}(3x)+ \frac{{\rm d}}{{\rm d}x} \left( y\frac{{\rm d}y}{{\rm d}x} \right) &=0 \\ &\Rightarrow& 3 + \underbrace{\left( \frac{{\rm d}}{{\rm d}x}y \right)\frac{{\rm d}y}{{\rm d}x} + y\frac{{\rm d}^2y}{{\rm d}x^2}}_{\rm product\;rule}&=0 \qquad{\rm by\;product\;rule} \\ &\Rightarrow& 3 + \left( \frac{{\rm d}y}{{\rm d}x} \right)^2 + y\frac{{\rm d}^2y}{{\rm d}x^2}&=0 \\ &\Rightarrow& 3 + \left( \frac{{\rm d}y}{{\rm d}x} \right)^2 + y\frac{{\rm d}^2y}{{\rm d}x^2}&=0 \\ &\Rightarrow& y\frac{{\rm d}^2y}{{\rm d}x^2}&=-\left[ 3 + \left( \frac{{\rm d}y}{{\rm d}x} \right)^2 \right] \\ &&&=-\left[ 3 + \left( -\frac{3x}{y} \right)^2 \right] \\ &&&=-\left( 3 + \frac{9x^2}{y^2} \right) \\ &&&=-\frac{9x^2+3y^2}{y^2} \\ &&&=-\frac{3a^2}{y^2} \\ \\ && \Rightarrow\qquad \frac{{\rm d}^2y}{{\rm d}x^2} &=-\frac{3a^2}{y^3}. \qquad\square \end{align} $$

 

Question 4. Given $x^3+y^3=a$, show that $$ \begin{align} \frac{{\rm d}y}{{\rm d}x}&=-\frac{x^2}{y^2} \\ \\ \frac{{\rm d}^2y}{{\rm d}x^2}&= - \frac{ 2ax }{ y^5 } \end{align} $$

 

Solution.

We use implicit differentiation which is an application of chain rule and product rule as we will see here.

$$ \begin{align} && x^3 + y^3 &= a \\ &\Rightarrow& \frac{{\rm d}}{{\rm d}x} \left( x^3 + y^3 \right) &=\frac{{\rm d}}{{\rm d}x} a \\ &\Rightarrow& \frac{{\rm d}}{{\rm d}x} x^3 + \frac{{\rm d}}{{\rm d}x} y^3 &=0 \\ &\Rightarrow& 3x^2 + \underbrace{ \frac{{\rm d}y}{{\rm d}x} \frac{{\rm d}}{{\rm d}y} }_{ {\rm chain\;rule} } y^3 &=0 \qquad{\rm by\;chain\;rule} \\ &\Rightarrow& 3x^2 + 3y^2 \frac{{\rm d}y}{{\rm d}x} &= 0 \\ &\Rightarrow& 3x^2 + 3y^2 \frac{{\rm d}y}{{\rm d}x} &= 0 \\ &\Rightarrow& \frac{{\rm d}y}{{\rm d}x} &= - \frac{ x^2 }{ y^2 }. \qquad\square \end{align} $$

 

For the second derivative, there are two different ways to find it.

 

(i) Method 1: By quotient rule. $$ \begin{align} && \frac{{\rm d}}{{\rm d}x}\left( \frac{u}{v} \right) &= \frac{u'v-uv'}{v^2} \\ \\ &\Rightarrow& \frac{{\rm d}^2y}{{\rm d}x^2} = \frac{{\rm d}}{{\rm d}x}\left( - \frac{ x^2 }{ y^2 } \right) &= - \frac{ \left(\frac{{\rm d}}{{\rm d}x} x^2 \right) y^2 - x^2 \left( \frac{{\rm d}}{{\rm d}x} y^2 \right) }{ y^4 } \\ &&&= - \frac{ 2xy^2 - 2x^2y \frac{{\rm d}y}{{\rm d}x} }{ y^4 } \\ &&&= - \frac{ 2xy \left[ y - x \left( - \frac{ x^2 }{ y^2 } \right) \right] }{ y^4 } \\ &&&= - \frac{ 2x \left( y^3 + x^3 \right) }{ y^5 } \\ &&&= - \frac{ 2ax }{ y^5 }. \qquad\square \end{align} $$

 

(ii) Method 2: By implicit differentiation. We start with: $$ \begin{align} && x^2 + y^2 \frac{{\rm d}y}{{\rm d}x} &= 0 \\ &\Rightarrow& \frac{{\rm d}}{{\rm d}x} x^2 + \frac{{\rm d}}{{\rm d}x} \left( y^2 \frac{{\rm d}y}{{\rm d}x} \right) &=0 \\ &\Rightarrow& 2x + \underbrace{ \left( \frac{{\rm d}}{{\rm d}x} y^2 \right)\frac{{\rm d}y}{{\rm d}x} + y^2 \frac{{\rm d}^2y}{{\rm d}x^2} }_{\rm product\;rule}&=0 \qquad{\rm by\;product\;rule} \\ &\Rightarrow& 2x + 2y \left( \frac{{\rm d}y}{{\rm d}x} \right)^2 + y^2 \frac{{\rm d}^2y}{{\rm d}x^2}&=0 \\ \\ &\Rightarrow& y^2 \frac{{\rm d}^2y}{{\rm d}x^2} &= - 2 \left[ x + y \left( \frac{{\rm d}y}{{\rm d}x} \right)^2 \right] \\ &&&= - 2 \left[ x + y \left( -\frac{ x^2 }{ y^2 } \right)^2 \right] \\ &&&= - 2 \left( x + \frac{ x^4 }{ y^3 } \right) \\ &&&= - 2 \frac{ x \left( x^3 + y^3 \right) }{ y^3 } \\ &&&= - \frac{ 2ax }{ y^3 } \\ \\ && \Rightarrow\qquad \frac{{\rm d}^2y}{{\rm d}x^2} &= - \frac{ 2ax }{ y^5 }. \qquad\square \end{align} $$

 

Question 5. Given $8x^3+xy^2=2y^4$, show that $$ \begin{align} \frac{{\rm d}y}{{\rm d}x}&=\frac{24x^2 + y^2}{2y \left( 4y^2 - x \right)} \\ \\ \frac{{\rm d}^2y}{{\rm d}x^2}&=\frac{ 8y^4 \left( 96x + 1 \right) \left( 4y^2-x \right) + 2\left( x - 12y^2 \right) \left( 24x^2 + y^2 \right)^2 }{ 8y^3\left( 4y^2-x \right)^3 } \end{align} $$

 

Solution.

We use implicit differentiation which is an application of chain rule and product rule as we will see here.

$$ \begin{align} && 8x^3+xy^2&=2y^4 \\ &\Rightarrow& \frac{{\rm d}}{{\rm d}x} \left( 8x^3 \right) + \frac{{\rm d}}{{\rm d}x} \left( xy^2 \right) &=\frac{{\rm d}}{{\rm d}x}\left(2y^4\right) \\ &\Rightarrow& \frac{{\rm d}}{{\rm d}x}\left(8x^3\right) + \underbrace{\left(\frac{{\rm d}}{{\rm d}x}x\right) y^2+x\left(\frac{{\rm d}}{{\rm d}x}y^2 \right)}_{\rm product\;rule} &=\frac{{\rm d}}{{\rm d}x}\left(2y^4\right) \\ &\Rightarrow& 24x^2 + y^2 + x\bigg(\underbrace{\frac{{\rm d}y}{{\rm d}x}\frac{{\rm d}}{{\rm d}y}}_{\rm chain\;rule}y^2 \bigg)&=\underbrace{\frac{{\rm d}y}{{\rm d}x}\frac{{\rm d}}{{\rm d}y}}_{\rm chain\;rule}2y^4 \\ &\Rightarrow& 24x^2 + y^2 + 2xy\frac{{\rm d}y}{{\rm d}x}&=8y^3\frac{{\rm d}y}{{\rm d}x} \\ &\Rightarrow& \left( 8y^3-2xy \right)\frac{{\rm d}y}{{\rm d}x} &=24x^2 + y^2 \\ &\Rightarrow& \frac{{\rm d}y}{{\rm d}x} &=\frac{24x^2 + y^2}{8y^3-2xy} =\frac{24x^2 + y^2}{2y \left( 4y^2 - x \right)} . \qquad\square \end{align} $$

 

For the second derivative, there are two different ways to find it.

 

(i) Method 1: By quotient rule. $$ \begin{align} \frac{{\rm d}}{{\rm d}x}\left( \frac{u}{v} \right)&=\frac{u'v-uv'}{v^2} \\ \\ \frac{{\rm d}^2y}{{\rm d}x^2}=\frac{{\rm d}}{{\rm d}x}\left( \frac{24x^2 + y^2}{8y^3-2xy} \right) &=\frac{\left( 8y^3-2xy \right)\frac{{\rm d}}{{\rm d}x}\left( 24x^2 + y^2 \right) - \left( 24x^2 + y^2 \right)\frac{{\rm d}}{{\rm d}x}\left( 8y^3-2xy \right)}{ \left( 8y^3-2xy \right)^2 } \\ &=\frac{\left( 8y^3-2xy \right)\left( 48x + \frac{{\rm d}y}{{\rm d}x}\frac{{\rm d}}{{\rm d}y}y^2 \right) - \left( 24x^2 + y^2 \right)\left( \frac{{\rm d}y}{{\rm d}x}\frac{{\rm d}}{{\rm d}y}8y^3-2y-2x\frac{{\rm d}y}{{\rm d}x} \right)}{ \left( 8y^3-2xy \right)^2 } \\ &=\frac{ \left( 8y^3-2xy \right) \left( 48x + 2y\frac{{\rm d}y}{{\rm d}x} \right) - \left( 24x^2 + y^2 \right) \left( 24y^2\frac{{\rm d}y}{{\rm d}x}-2y-2x\frac{{\rm d}y}{{\rm d}x} \right) }{ \left( 8y^3-2xy \right)^2 } \\ &=\frac{ 48x \left( 8y^3-2xy \right) + 2y \left( 24x^2 + y^2 \right) + \left[ 2y\left( 8y^3-2xy \right) - \left( 24y^2 - 2x \right) \left( 24x^2 + y^2 \right) \right] \frac{{\rm d}y}{{\rm d}x} }{ \left( 8y^3-2xy \right)^2 } \\ &=\frac{ 48x \left( 8y^3-2xy \right) + 2y \left( 24x^2 + y^2 \right) + \left[ 2y\left( 8y^3-2xy \right) - \left( 24y^2 - 2x \right) \left( 24x^2 + y^2 \right) \right] \frac{24x^2 + y^2}{8y^3-2xy} }{ \left( 8y^3-2xy \right)^2 } \\ &=\frac{ 48x \left( 8y^3-2xy \right) + 2y \left( 24x^2 + y^2 \right) + 2y\left( 24x^2 + y^2 \right) - \left( 24y^2 - 2x \right) \left( 24x^2 + y^2 \right) \frac{24x^2 + y^2}{8y^3-2xy} }{ \left( 8y^3-2xy \right)^2 } \\ &=\frac{ 96xy \left( 4y^2-x \right) + 4y \left( 24x^2 + y^2 \right) - \frac{ \left( 24y^2 - 2x \right) \left( 24x^2 + y^2 \right)^2 }{8y^3-2xy} }{ \left( 8y^3-2xy \right)^2 } \\ &=\frac{ 4y \left[ 24x\left( 4y^2-x \right) + \left( 24x^2 + y^2 \right) \right] \left( 8y^3-2xy \right) - \left( 24y^2 - 2x \right) \left( 24x^2 + y^2 \right)^2 }{ \left( 8y^3-2xy \right)^2 } \\ &=\frac{ 4y \left( 96xy^2 + y^2 \right) \left( 8y^3-2xy \right) - \left( 24y^2 - 2x \right) \left( 24x^2 + y^2 \right)^2 }{ \left( 8y^3-2xy \right)^2 } \\ &=\frac{ 8y^4 \left( 96x + 1 \right) \left( 4y^2-x \right) + 2\left( x - 12y^2 \right) \left( 24x^2 + y^2 \right)^2 }{ 8y^3\left( 4y^2-x \right)^2 }.\qquad\square \end{align} $$ Aside: In order to simplify the expression further, we may look at the original equation or its variations $$ \begin{align} && 8x^3+xy^2=2y^4 \\ &\Leftrightarrow& x(8x+y^2)=2y^4 \\ &\Leftrightarrow& y^2(2y^2-x)=8x^3 \end{align} $$ but none of them seems directly helpful...

 

(ii) Method 2: By implicit differentiation. We start with: $$ \begin{align} && \left( 8y^3-2xy \right)\frac{{\rm d}y}{{\rm d}x} &=24x^2 + y^2 \\ &\Rightarrow& \frac{{\rm d}}{{\rm d}x}\left[ \left( 8y^3-2xy \right)\frac{{\rm d}y}{{\rm d}x} \right] &= \frac{{\rm d}}{{\rm d}x}\left( 24x^2 + y^2 \right) \\ &\Rightarrow& \underbrace{\frac{{\rm d}}{{\rm d}x}\left( 8y^3-2xy \right)\left(\frac{{\rm d}y}{{\rm d}x}\right)+\left( 8y^3-2xy \right)\frac{{\rm d}^2y}{{\rm d}x^2}}_{\rm product\;rule} &= \frac{{\rm d}}{{\rm d}x}\left( 24x^2 \right) + \frac{{\rm d}}{{\rm d}x} y^2 \\ &\Rightarrow& \left[ \frac{{\rm d}}{{\rm d}x}\left(8y^3\right) - \frac{{\rm d}}{{\rm d}x}(2xy) \right]\left(\frac{{\rm d}y}{{\rm d}x}\right) + \left( 8y^3-2xy \right)\frac{{\rm d}^2y}{{\rm d}x^2} &= 48x + \frac{{\rm d}}{{\rm d}x} y^2 \\ &\Rightarrow& \bigg[ \underbrace{\frac{{\rm d}y}{{\rm d}x}\frac{{\rm d}}{{\rm d}y}}_{\rm chain\;rule}8y^3 - \underbrace{\left( \frac{{\rm d}}{{\rm d}x}(2x)y + 2x\frac{{\rm d}y}{{\rm d}x} \right)}_{\rm product\;rule} \bigg] \left( \frac{{\rm d}y}{{\rm d}x} \right) + \left( 8y^3-2xy \right)\frac{{\rm d}^2y}{{\rm d}x^2} &= 48x + \underbrace{\frac{{\rm d}y}{{\rm d}x}\frac{{\rm d}}{{\rm d}y}}_{\rm chain\;rule} y^2 \\ &\Rightarrow& \left( 24y^2\frac{{\rm d}y}{{\rm d}x} - 2y - 2x \frac{{\rm d}y}{{\rm d}x} \right)\left( \frac{{\rm d}y}{{\rm d}x} \right) + \left( 8y^3-2xy \right)\frac{{\rm d}^2y}{{\rm d}x^2} &= 48x + 2y\frac{{\rm d}y}{{\rm d}x} \\ &\Rightarrow& 24y^2\left( \frac{{\rm d}y}{{\rm d}x} \right)^2 - 2y \frac{{\rm d}y}{{\rm d}x} - 2x\left(\frac{{\rm d}y}{{\rm d}x}\right)^2 + \left( 8y^3-2xy \right)\frac{{\rm d}^2y}{{\rm d}x^2} &= 48x + 2y\frac{{\rm d}y}{{\rm d}x} \end{align} $$ which gives $$ \begin{align} &\Rightarrow& \left( 8y^3-2xy \right)\frac{{\rm d}^2y}{{\rm d}x^2} &= 48x + 4y\frac{{\rm d}y}{{\rm d}x} + \left( 2x - 24y^2 \right)\left(\frac{{\rm d}y}{{\rm d}x}\right)^2 \\ &\Rightarrow& \frac{{\rm d}^2y}{{\rm d}x^2} &=\frac{ 48x + 4y\frac{{\rm d}y}{{\rm d}x} + \left( 2x - 24y^2 \right)\left(\frac{{\rm d}y}{{\rm d}x}\right)^2 }{ \left( 8y^3-2xy \right) } \\ &&&=\frac{ 48x + 4y\left( \frac{24x^2 + y^2}{8y^3-2xy} \right) + \left( 2x - 24y^2 \right)\left( \frac{24x^2 + y^2}{8y^3-2xy} \right)^2 }{ \left( 8y^3-2xy \right) } \\ &&&=\frac{ 48x\left( 8y^3-2xy \right)^2 + 4y\left( 24x^2 + y^2 \right) \left( 8y^3-2xy \right) + \left( 2x - 24y^2 \right) \left( 24x^2 + y^2 \right)^2 }{ \left( 8y^3-2xy \right)^3 } \\ &&&=\frac{ 4 \left[ 12x\left( 8y^3-2xy \right) + y\left( 24x^2 + y^2 \right) \right] \left( 8y^3-2xy \right) + \left( 2x - 24y^2 \right) \left( 24x^2 + y^2 \right)^2 }{ \left( 8y^3-2xy \right)^3 } \\ &&&=\frac{ 4 \left( 96xy^3 - 24x^2y + 24x^2y + y^3 \right) \left( 8y^3-2xy \right) + \left( 2x - 24y^2 \right) \left( 24x^2 + y^2 \right)^2 }{ \left( 8y^3-2xy \right)^3 } \\ &&&=\frac{ 4y^3 \left( 96x + 1 \right) \left( 8y^3-2xy \right) + \left( 2x - 24y^2 \right) \left( 24x^2 + y^2 \right)^2 }{ \left( 8y^3-2xy \right)^3 } \\ &&&=\frac{ 8y^4 \left( 96x + 1 \right) \left( 4y^2-x \right) + 2\left( x - 12y^2 \right) \left( 24x^2 + y^2 \right)^2 }{ 8y^3\left( 4y^2-x \right)^3 }.\qquad\square \end{align} $$

 

Question 6. Given $x+y={\rm e}^{x-y}$, show that $$ \begin{align} \frac{{\rm d}y}{{\rm d}x}&= \frac{ x + y - 1 }{ x + y + 1 } \\ \\ \frac{{\rm d}^2y}{{\rm d}x^2}&=-\frac{2x}{5y^5} \end{align} $$

 

Solution.

We use implicit differentiation which is an application of chain rule and product rule as we will see here.

$$ \begin{align} && x + y &= \textrm e^x \textrm e^{-y} \\ &\Rightarrow& \frac{{\rm d}}{{\rm d}x} \left( x + y \right) &=\frac{{\rm d}}{{\rm d}x} \left( \textrm e^x \textrm e^{-y} \right) \\ &\Rightarrow& \frac{{\rm d}}{{\rm d}x} x + \frac{{\rm d}}{{\rm d}x} y &= \textrm e^{-y} \frac{{\rm d}}{{\rm d}x} \textrm e^x + \textrm e^x \underbrace{ \frac{{\rm d}y}{{\rm d}x} \frac{{\rm d}}{{\rm d}y} }_{ {\rm chain\;rule} } \textrm e^{-y} \qquad{\rm by\;chain\;rule} \\ &\Rightarrow& 1 + \frac{{\rm d} y}{{\rm d}x} &= \textrm e^{-y} \textrm e^{x} - \textrm e^{x} \textrm e^{-y} \frac{{\rm d}y}{{\rm d}x} \\ &&&= \textrm e^{x-y} \left( 1 + \frac{{\rm d}y}{{\rm d}x} \right) \\ \\ &\Rightarrow& \left( 1 - \textrm e^{x-y} \right) \frac{{\rm d}y}{{\rm d}x} &= \textrm e^{x-y} - 1 \\ \\ &\Rightarrow& \frac{{\rm d}y}{{\rm d}x} &= \frac{ \textrm e^{x-y} - 1 }{ \textrm e^{x-y} + 1 } = \frac{ \textrm e^{x} - \textrm e^{y} }{ \textrm e^{x} + \textrm e^{y} } = \frac{ x + y - 1 }{ x + y + 1 }. \qquad\square \end{align} $$

 

For the second derivative, there are two different ways to find it.

 

(i) Method 1: By quotient rule. $$ \begin{align} && \frac{{\rm d}}{{\rm d}x}\left( \frac{u}{v} \right) &= \frac{u'v-uv'}{v^2} \\ \\ &\Rightarrow& \frac{{\rm d}^2y}{{\rm d}x^2} = \frac{{\rm d}}{{\rm d}x} \left( \frac{ x + y - 1 }{ x + y + 1 } \right) &= - \frac{ \left[ \frac{{\rm d}}{{\rm d}x} \left( x + y - 1 \right) \right] \left( x + y + 1 \right) - \left( x + y - 1 \right) \left[ \frac{{\rm d}}{{\rm d}x} \left( x + y + 1 \right) \right] }{ \left( x + y + 1 \right)^2 } \\ &&&= - \frac{ \left( 1 + \frac{{\rm d}y}{{\rm d}x} \right) \left( x + y + 1 \right) - \left( x + y - 1 \right) \left( 1 + \frac{{\rm d}y}{{\rm d}x} \right) }{ \left( x + y + 1 \right)^2 } \\ &&&= - \frac{ 2 \left( 1 + \frac{{\rm d}y}{{\rm d}x} \right) }{ \left( x + y + 1 \right)^2 } \\ &&&= - \frac{ 2 \left( 1 + \frac{ x + y - 1 }{ x + y + 1 } \right) }{ \left( x + y + 1 \right)^2 } \\ &&&= - \frac{ 4 \left( x + y \right) }{ \left( x + y + 1 \right)^3 }. \qquad\square \end{align} $$

 

(ii) Method 2: By implicit differentiation. We start with: $$ \begin{align} && \left( x + y + 1 \right) \frac{{\rm d}y}{{\rm d}x} &= x + y - 1 \\ &\Rightarrow& \left[ \frac{{\rm d}}{{\rm d}x} \left( x + y + 1 \right) \right] \frac{{\rm d}y}{{\rm d}x} + \left( x + y + 1 \right) \frac{{\rm d}^2y}{{\rm d}x^2} &=\frac{{\rm d}}{{\rm d}x} \left( x + y - 1 \right) \qquad{\rm by\;product\;rule} \\ &\Rightarrow& \left( 1 + \frac{{\rm d}y}{{\rm d}x} \right)\frac{{\rm d}y}{{\rm d}x} + \left( x + y + 1 \right) \frac{{\rm d}^2y}{{\rm d}x^2} &= 1 + \frac{{\rm d}y}{{\rm d}x} \\ \\ &\Rightarrow& \left( x + y + 1 \right) \frac{{\rm d}^2y}{{\rm d}x^2} &= \left( 1 + \frac{{\rm d}y}{{\rm d}x} \right) \left( \frac{{\rm d}y}{{\rm d}x} - 1 \right) \\ &= \left( 1 + \frac{ x + y - 1 }{ x + y + 1 } \right) \left( \frac{ x + y - 1 }{ x + y + 1 } - 1 \right) \\ &= \frac{ 2(x+y)\cdot 2 }{ (x + y + 1)^2 } \\ &= \frac{ 4(x+y) }{ (x + y + 1)^2 } \\ \\ &\Rightarrow& \frac{{\rm d}^2y}{{\rm d}x^2} &= \frac{ 4(x+y) }{ (x + y + 1)^3 }. \qquad\square \end{align} $$

 

Aside. Alternatively, one could derive $ \frac{{\rm d}^2y}{{\rm d}x^2} $ using the exponential expression, $\left( \textrm e^{x-y} + 1 \right) \frac{{\rm d}y}{{\rm d}x} = \left( \textrm e^{x-y} - 1 \right)$

 

(i) Method 1: By quotient rule. $$ \begin{align} && \frac{{\rm d}}{{\rm d}x}\left( \frac{u}{v} \right) &= \frac{u'v-uv'}{v^2} \\ \\ &\Rightarrow& \frac{{\rm d}^2y}{{\rm d}x^2} = \frac{{\rm d}}{{\rm d}x} \left( \frac{ \textrm e^{x} - \textrm e^{y} }{ \textrm e^{x} + \textrm e^{y} } \right) &= - \frac{ \left[ \frac{{\rm d}}{{\rm d}x} \left( \textrm e^{x} - \textrm e^{y} \right) \right] \left( \textrm e^{x} + \textrm e^{y} \right) - \left( \textrm e^{x} - \textrm e^{y} \right) \left[ \frac{{\rm d}}{{\rm d}x} \left( \textrm e^{x} + \textrm e^{y} \right) \right] }{ \left( \textrm e^{x} + \textrm e^{y} \right)^2 } \\ &&&= - \frac{ \left( \textrm e^x - \textrm e^y \frac{{\rm d}y}{{\rm d}x} \right) \left( \textrm e^x + \textrm e^y \right) - \left( \textrm e^x - \textrm e^y \right) \left( \textrm e^x + \textrm e^y \frac{{\rm d}y}{{\rm d}x} \right) }{ \left( \textrm e^{x} + \textrm e^{y} \right)^2 } \\ &&&= - \frac{ \textrm e^x \left( \textrm e^x + \textrm e^y \right) - \left( \textrm e^x + \textrm e^y \right) \textrm e^y \frac{{\rm d}y}{{\rm d}x} - \left( \textrm e^x - \textrm e^y \right) \textrm e^x - \left( \textrm e^x - \textrm e^y \right) \textrm e^y \frac{{\rm d}y}{{\rm d}x} }{ \left( \textrm e^{x} + \textrm e^{y} \right)^2 } \\ &&&= - \frac{ 2 \textrm e^{ x + y } - 2 \textrm e^{ x + y } \frac{{\rm d}y}{{\rm d}x} }{ \left( \textrm e^{x} + \textrm e^{y} \right)^2 } \\ &&&= - \left( 1 - \frac{{\rm d}y}{{\rm d}x} \right) \frac{ 2 \textrm e^{ x + y } }{ \left( \textrm e^{x} + \textrm e^{y} \right)^2 } \\ &&&= - \left( 1 - \frac{ \textrm e^{x} - \textrm e^{y} }{ \textrm e^{x} + \textrm e^{y} } \right) \frac{ 2 \textrm e^{ x + y } }{ \left( \textrm e^{x} + \textrm e^{y} \right)^2 } \\ &&&= - \left[ \frac{ \left( \textrm e^{x} + \textrm e^{y} \right) - \left( \textrm e^{x} - \textrm e^{y} \right) }{ \textrm e^{x} + \textrm e^{y} } \right] \frac{ 2 \textrm e^{ x + y } }{ \left( \textrm e^{x} + \textrm e^{y} \right)^2 } \\ &&&= - \left[ \frac{ 2 \textrm e^{y} }{ \textrm e^{x} + \textrm e^{y} } \right] \frac{ 2 \textrm e^{ x + y } }{ \left( \textrm e^{x} + \textrm e^{y} \right)^2 } \\ &&&= - \frac{ 4 \textrm e^{ x + 2y } }{ \left( \textrm e^{x} + \textrm e^{y} \right)^3 } \\ &&&= - \frac{ 4 \textrm e^{ x - y } }{ \left( \textrm e^{x - y} + 1 \right)^3 } \\ &&&= - \frac{ 4 (x+y) }{ \left( x + y + 1 \right)^3 }. \qquad\square \end{align} $$

 

(ii) Method 2: By implicit differentiation. We start with: $$ \begin{align} && \left( \textrm e^{x-y} + 1 \right) \frac{{\rm d}y}{{\rm d}x} &= \textrm e^{x-y} - 1 \\ &\Rightarrow& \frac{{\rm d}}{{\rm d}x} \left( \textrm e^{x-y} + 1 \right) \frac{{\rm d}y}{{\rm d}x} + \left( \textrm e^{x-y} + 1 \right) \frac{{\rm d}^2y}{{\rm d}x^2} &=\frac{{\rm d}}{{\rm d}x} \left( \textrm e^{x-y} - 1 \right) \\ &\Rightarrow& \textrm e^{x-y} \left( 1 - \frac{{\rm d}y}{{\rm d}x} \right) \frac{{\rm d}y}{{\rm d}x} + \left( \textrm e^{x-y} + 1 \right) \frac{{\rm d}^2y}{{\rm d}x^2} &= \textrm e^{x-y} \left( 1 - \frac{{\rm d}y}{{\rm d}x} \right) \\ \\ &\Rightarrow& \left( \textrm e^{x-y} + 1 \right) \frac{{\rm d}^2y}{{\rm d}x^2} &= \textrm e^{x-y} \left( 1 - \frac{{\rm d}y}{{\rm d}x} \right) \left( \frac{{\rm d}y}{{\rm d}x} - 1 \right) \\ &&&= \textrm e^{x-y} \left( 1 - \frac{ \textrm e^{x} - \textrm e^{y} }{ \textrm e^{x} + \textrm e^{y} } \right) \left( \frac{ \textrm e^{x} - \textrm e^{y} }{ \textrm e^{x} + \textrm e^{y} } - 1 \right) \\ &&&= \textrm e^{x-y} \left( \frac{ 2 \textrm e^{y} }{ \textrm e^{x} + \textrm e^{y} } \right) \left( \frac{ - 2 \textrm e^{y} }{ \textrm e^{x} + \textrm e^{y} } \right) \\ &&&= -4 \textrm e^{x-y} \left[ \frac{ \textrm e^{2y} }{ \left( \textrm e^{x} + \textrm e^{y} \right)^2 } \right] \\ &&&= - \frac{ 4 \textrm e^{ x - y } }{ \left( \textrm e^{ x - y } + 1 \right)^2 } \\ \\ &\Rightarrow& \frac{{\rm d}^2y}{{\rm d}x^2} &= - \frac{ 4 \textrm e^{ x - y } }{ \left( \textrm e^{ x - y } + 1 \right)^3 } \\ &= - \frac{ 4 (x+y) }{ \left( x + y + 1 \right)^3 }. \qquad\square \end{align} $$

 

Question 7. Given $x^n+y^n=1$, show that $$ \begin{align} \frac{{\rm d}y}{{\rm d}x}&=-\frac{x^{n-1}}{y^{n-1}} \\ \\ \frac{{\rm d}^2y}{{\rm d}x^2}&=-\frac{(n-1)x^{n-2}}{y^{2n-1}} \end{align} $$

 

Solution.

We use implicit differentiation which is an application of chain rule and product rule as we will see here.

$$ \begin{align} && x^n+y^n&=1 \\ &\Rightarrow& \frac{{\rm d}}{{\rm d}x} \left( x^n+y^n \right) &=\frac{{\rm d}}{{\rm d}x}1 \\ &\Rightarrow& \frac{{\rm d}}{{\rm d}x}x^n + \frac{{\rm d}}{{\rm d}x}y^n &=0 \\ &\Rightarrow& nx^{n-1}+\underbrace{\frac{{\rm d}y}{{\rm d}x}\frac{{\rm d}}{{\rm d}y}}_{{\rm chain\;rule}}y^n&=0 \qquad{\rm by\;chain\;rule} \\ &\Rightarrow& nx^{n-1}+ny^{n-1}\frac{{\rm d}y}{{\rm d}x}&=0 \\ &\Rightarrow& x^{n-1}+y^{n-1}\frac{{\rm d}y}{{\rm d}x}&=0\qquad{\rm since}\quad n\ne0 \\ &\Rightarrow& \frac{{\rm d}y}{{\rm d}x}&=-\frac{x^{n-1}}{y^{n-1}} \end{align} $$

 

For the second derivative, there are two different ways to find it.

 

(i) Method 1: By quotient rule. $$ \begin{align} && \frac{{\rm d}}{{\rm d}x}\left( \frac{u}{v} \right)&=\frac{u'v-uv'}{v^2} \\ \\ &\Rightarrow& \frac{{\rm d}^2y}{{\rm d}x^2}=\frac{{\rm d}}{{\rm d}x}\left( -\frac{x^{n-1}}{y^{n-1}} \right) &=-\frac{\left(\frac{{\rm d}}{{\rm d}x}x^{n-1}\right)y^{n-1} - x^{n-1}\left( \frac{{\rm d}}{{\rm d}x}y^{n-1}\right)}{\left(y^{n-}\right)^2} \\ &&&=-\frac{(n-1)x^{n-2}y^{n-1} - x^{n-1}\bigg[ \overbrace{\frac{{\rm d}y}{{\rm d}x}\frac{{\rm d}}{{\rm d}y}}^{\rm chain\;rule}y^{n-1} \bigg] }{y^{2n-2}} \\ &&&=-\frac{(n-1)x^{n-2}y^{n-1} - x^{n-1}\left[ (n-1)y^{n-2}\frac{{\rm d}y}{{\rm d}x} \right]}{y^{2n-2}} \\ &&&=-\frac{(n-1)x^{n-2}y^{n-1} - (n-1)x^{n-1}y^{n-2}\frac{{\rm d}y}{{\rm d}x}}{y^{2n-2}} \\ &&&=-\frac{(n-1)x^{n-2}y^{n-1} \left( 1 - \frac{x}{y}\frac{{\rm d}y}{{\rm d}x} \right) }{y^{2n-2}} \\ &&&=-\frac{(n-1)x^{n-2}y^{n-1} \left[ 1 - \frac{x}{y}\left( -\frac{x^{n-1}}{y^{n-1}} \right) \right] }{y^{2n-2}} \\ &&&=-\frac{(n-1)x^{n-2}y^{n-1} \left( 1 + \frac{x^n}{y^n} \right) }{y^{2n-2}} \\ &&&=-\frac{(n-1)x^{n-2}y^{n-1} \left( \frac{y^n + x^n }{y^n} \right) }{y^{2n-2}} \\ &&&=-\frac{(n-1)x^{n-2}\frac1y }{y^{2n-2}} \qquad{\rm since}\quad x^n+y^n=1 \\ &&&=-\frac{(n-1)x^{n-2}}{y^{2n-1}} .\qquad\square \end{align} $$

 

(ii) Method 2: By implicit differentiation. We start with: $$ \begin{align} && x^{n-1}+y^{n-1}\frac{{\rm d}y}{{\rm d}x}&=0 \\ &\Rightarrow& \frac{{\rm d}}{{\rm d}x} x^{n-1}+ \frac{{\rm d}}{{\rm d}x} \left( y^{n-1}\frac{{\rm d}y}{{\rm d}x} \right) &=0 \\ &\Rightarrow& (n-1)x^{n-2}+ \underbrace{\left( \frac{{\rm d}}{{\rm d}x}y^{n-1} \right) + y^{n-1}\frac{{\rm d}^2y}{{\rm d}x^2}}_{\rm product\;rule}&=0 \qquad{\rm by\;product\;rule} \\ &\Rightarrow& (n-1)x^{n-2}+ \underbrace{\frac{{\rm d}y}{{\rm d}x}\frac{{\rm d}}{{\rm d}y}}_{{\rm chain\;rule}} y^{n-1} + y^{n-1}\frac{{\rm d}^2y}{{\rm d}x^2}&=0 \qquad{\rm by\;chain\;rule} \\ &\Rightarrow& (n-1)x^{n-2} + (n-1)y^{n-2}\left( \frac{{\rm d}y}{{\rm d}x} \right)^2 + y^{n-1}\frac{{\rm d}^2y}{{\rm d}x^2}&=0 \\ &\Rightarrow& y^{n-1}\frac{{\rm d}^2y}{{\rm d}x^2}&=-(n-1)\left[ x^{n-2} + y^{n-2} \left( \frac{{\rm d}y}{{\rm d}x} \right)^2 \right] \\ &&&=-(n-1)\left[ x^{n-2} + y^{n-2}\left(-\frac{x^{n-1}}{y^{n-1}}\right)^2 \right] \\ &&&=-(n-1)\left( x^{n-2} + y^{n-2} \frac{x^{2n-2}}{y^{2n-2}} \right) \\ &&&=-(n-1)\left( x^{n-2} + \frac{x^{2n-2}}{y^n} \right) \\ &&&=-(n-1)x^{n-2}\left( 1 + \frac{x^n}{y^n} \right) \\ &&&=-(n-1)x^{n-2}\left( \frac{x^n+y^n}{y^n} \right) \\ &&&=-\frac{(n-1)x^{n-2}}{y^n} \qquad{\rm since}\quad x^n+y^n=1 \\ \\ && \Rightarrow\qquad \frac{{\rm d}^2y}{{\rm d}x^2} &=-\frac{(n-1)x^{n-2}}{y^{2n-1}}.\qquad\square \end{align} $$

 

Question 8. Given $ a x^m + b y^n = c r^k $, show that $$ \begin{align} \frac{{\rm d}y}{{\rm d}x }&= - \left( \frac{ a m }{ b n } \right) \frac{ x^{m-1} }{ y^{n-1} } \\ \\ \frac{{\rm d}^2y}{{\rm d}x^2}&= - \left( \frac{ a m }{ b n } \right) \left[ (m-1) y^n + \left( \frac{ am }{ bn } \right) (n-1) x^m \right] \frac{ x^{ m - 2 } }{ y^{ 2n - 1 } } \end{align} $$

 

Solution.

We use implicit differentiation which is an application of chain rule and product rule as we will see here.

$$ \begin{align} && a x^m + b y^n &= c r^k \\ &\Rightarrow& \frac{{\rm d}}{{\rm d}x} \left( a x^m + b y^n \right) &=\frac{{\rm d}}{{\rm d}x} \left( c r^k \right) \\ &\Rightarrow& a \frac{{\rm d}}{{\rm d}x} x^m + b \frac{{\rm d}}{{\rm d}x} y^n &=0 \\ &\Rightarrow& am x^{m-1} + b \underbrace{ \frac{{\rm d}y}{{\rm d}x} \frac{{\rm d}}{{\rm d}y} }_{ {\rm chain\;rule} } y^n &=0 \qquad{\rm by\;chain\;rule} \\ &\Rightarrow& am x^{m-1} + bn y^{n-1} \frac{{\rm d}y}{{\rm d}x} &= 0 \\ &\Rightarrow& am x^{m-1} + bn y^{n-1} \frac{{\rm d}y}{{\rm d}x} &= 0 \\ &\Rightarrow& \frac{{\rm d}y}{{\rm d}x} &= - \left( \frac{ a m }{ b n } \right) \frac{ x^{m-1} }{ y^{n-1} }. \qquad\square \end{align} $$

 

For the second derivative, there are two different ways to find it.

 

(i) Method 1: By quotient rule. $$ \begin{align} && \frac{{\rm d}}{{\rm d}x}\left( \frac{u}{v} \right) &= \frac{u'v-uv'}{v^2} \\ \\ &\Rightarrow& \frac{{\rm d}^2y}{{\rm d}x^2} = - \left( \frac{ a m }{ b n } \right) \frac{{\rm d}}{{\rm d}x} \left( \frac{ x^{m-1} }{ y^{n-1} } \right) &= - \left( \frac{ a m }{ b n } \right) \frac{ \left(\frac{{\rm d}}{{\rm d}x} x^{m-1} \right) y^{n-1} - x^{m-1} \left( \frac{{\rm d}}{{\rm d}x} y^{n-1} \right) }{ y^{2(n-1)} } \\ &&&= - \left( \frac{ a m }{ b n } \right) \frac{ (m-1) x^{m-2} y^{n-1} - (n-1) x^{m-1} y^{n-2} \frac{{\rm d}y}{{\rm d}x} }{ y^{2(n-1)} } \\ &&&= - \left( \frac{ a m }{ b n } \right) \frac{ \left[ (m-1) y - (n-1) x \left( - \frac{ a m }{ b n } \frac{ x^{m-1} }{ y^{n-1} } \right) \right] x^{m-2} y^{n-2} }{ y^{2(n-1)} } \\ &&&= - \left( \frac{ a m }{ b n } \right) \left[ (m-1) y^n + \left( \frac{ a m }{ b n } \right) (n-1) x^m \right] \frac{ x^{m-2} y^{n-2} }{ y^{n-1} y^{2(n-1)} } \\ &&&= - \left( \frac{ a m }{ b n } \right) \left[ (m-1) y^n + \left( \frac{ a m }{ b n } \right) (n-1) x^m \right] \frac{ x^{m-2} }{ y^{2n-1} }. \qquad\square \end{align} $$

 

(ii) Method 2: By implicit differentiation. We start with: $$ \begin{align} && am x^{m-1} + bn y^{n-1} \frac{{\rm d}y}{{\rm d}x} &= 0 \\ &\Rightarrow& am \frac{{\rm d}}{{\rm d}x} x^{m-1} + bn \frac{{\rm d}}{{\rm d}x} \left( y^{n-1} \frac{{\rm d}y}{{\rm d}x} \right) &=0 \\ &\Rightarrow& am(m-1)x^{m-2} + bn \underbrace{ \left( \frac{{\rm d}}{{\rm d}x} y^{n-1} \right) \frac{{\rm d}y}{{\rm d}x} + bn y^{n-1} \frac{{\rm d}^2y}{{\rm d}x^2} }_{\rm product\;rule}&=0 \qquad{\rm by\;product\;rule} \\ &\Rightarrow& am(m-1)x^{m-2} + bn(n-1) y^{n-2} \left( \frac{{\rm d}y}{{\rm d}x} \right)^2 + bn y^{n-1} \frac{{\rm d}^2y}{{\rm d}x^2}&=0 \\ \\ \end{align} $$ $$ \begin{align} &\Rightarrow& bn y^{n-1} \frac{{\rm d}^2y}{{\rm d}x^2} &= - \left[ am(m-1)x^{m-2} + bn(n-1) y^{n-2} \left( \frac{{\rm d}y}{{\rm d}x} \right)^2 \right] \\ &&&= - \left[ am(m-1)x^{m-2} + bn(n-1) y^{n-2} \left( - \frac{ a m }{ b n } \frac{ x^{m-1} }{ y^{n-1} } \right)^2 \right] \\ &&&= - \left[ am(m-1)x^{m-2} + bn(n-1) y^{n-2} \left( \frac{ a m }{ b n } \right)^2 \frac{ x^{2(m-1)} }{ y^{2(n-1)} } \right] \\ &&&= - \left[ am(m-1)x^{m-2} + bn(n-1) \left( \frac{ a m }{ b n } \right)^2 \frac{ x^{ 2(m-1) } }{ y^{ n } } \right] \\ &&&= - \left[ am(m-1) y^n + bn(n-1) \left( \frac{ a m }{ b n } \right)^2 x^m \right] \frac{ x^{ m - 2 } }{ y^{ n } } \\ &&&= - \left( \frac{ a m }{ b n } \right) \Big[ b n (m-1) y^n + a m (n-1) x^m \Big] \frac{ x^{ m - 2 } }{ y^{ n } } \\ \\ &\Rightarrow& \frac{{\rm d}^2y}{{\rm d}x^2} &= - \left( \frac{ a m }{ b n } \right) \left[ (m-1) y^n + \left( \frac{ am }{ bn } \right) (n-1) x^m \right] \frac{ x^{ m - 2 } }{ y^{ 2n - 1 } } \qquad \square \end{align} $$

 

Check. For $m=n$, i.e. $ax^n+by^n=cr^k$ $$ \begin{align} \frac{{\rm d}y}{{\rm d}x }&= - \left( \frac{ a }{ b } \right) \frac{ x^{n-1} }{ y^{n-1} } \\ \\ \frac{{\rm d}^2y}{{\rm d}x^2}&= - \left( \frac{ a }{ b } \right) (n-1) \left[ y^n + \left( \frac{ a}{ b} \right) x^n \right] \frac{ x^{ n - 2 } }{ y^{ 2n - 1 } } \end{align} $$ In addition, if $a=b$, i.e. $x^n+y^n= \frac{c}{a}r^k$ $$ \begin{align} \frac{{\rm d}y}{{\rm d}x }&= - \frac{ x^{n-1} }{ y^{n-1} } \qquad \checkmark \\ \\ \frac{{\rm d}^2y}{{\rm d}x^2}&= - (n-1) \left( x^n + y^n \right) \frac{ x^{ n - 2 } }{ y^{ 2n - 1 } } = - (n-1) \left( \frac{c}{a}r^k \right) \frac{ x^{ n - 2 } }{ y^{ 2n - 1 } } \qquad \checkmark \end{align} $$