7. Differentiation - Some inverse trigonometric functions

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Question. Find $\frac{dy}{dx}$ for: $$\begin{align} \textrm{(a)}&\qquad y = \arcsin(1-2x) \\ \textrm{(b)}&\qquad y = \arctan\left(x^2+1\right) \end{align}$$

 

Solution. (a) For $y=\arcsin(1-2x)$, we have $$\begin{gather} -1\le 1-2x \le 1 \\ -\frac{\pi}{2}\le y \le\frac{\pi}{2} \end{gather}$$ We re-express the equation $$\begin{align} y&=\arcsin(1-2x) \\ \Rightarrow\quad 1-2x&=\sin y. \end{align}$$ By implicit differentiation $$\begin{align} \frac{d}{dy}(1-2x)&=\frac{d}{dy}\sin y \\ \Rightarrow\quad -2\frac{dx}{dy}&=\cos y \\ &=\sqrt{1-\sin^2y}\qquad\left({\rm for}\quad -\frac{\pi}{2}\le y \le\frac{\pi}{2}\right) \\ &=\sqrt{1-(1-2x)^2} \\ &=\sqrt{4x-4x^2} \\ &=2\sqrt{x(1-x)} \\ \Rightarrow\quad \frac{dx}{dy}&=-\sqrt{x(1-x)} \\ \\ \Rightarrow\quad \frac{dy}{dx}&=\frac{1}{\frac{dx}{dy}} =-\frac{2}{\sqrt{1-(1-2x)^2}} =-\frac1{\sqrt{x(1-x)}} \end{align}$$

 

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(b) For $y=\arctan\left(x^2+1\right)$, we have $$\begin{gather} -\infty\le x^2+1 \le \infty \\ -\frac{\pi}{2}\le y \le\frac{\pi}{2} \end{gather}$$ We re-express the equation $$\begin{align} y&=\arctan\left(x^2+1\right) \\ \Rightarrow\quad x^2+1&=\tan y \end{align}$$ By chain rule, $$\begin{align} \frac{d}{dx}\left(x^2+1\right)&=\frac{d}{dx}\tan y \\ \Rightarrow\quad 2x&=\frac{dy}{dx}\frac{d}{dy}\tan y \\ &=\frac{dy}{dx}\sec^2 y \\ &=\frac{dy}{dx}\left(1+\tan^2y\right) \\ &=\frac{dy}{dx}\left[1+\left(x^2+1\right)^2\right] \\ \\ \Rightarrow\quad \frac{dy}{dx}&=\frac{2x}{1+\left(x^2+1\right)^2}=\frac{2x}{x^4+2x^2+2} \end{align}$$