Quiz 6 - More chain rule and differentiation of trigonometric functions

Problem 1

Question

Differentiate \(y = \sin^{-1}(e^{-x}) + \sec^{-1}(e^x)\).

Solution

\[ \begin{align} y &= \sin^{-1}(e^{-x}) + \sec^{-1}(e^x) \\ \frac{d}{dx}y &= \frac{d}{dx}\left(\sin^{-1}(e^{-x}) + \sec^{-1}(e^x)\right) \\ &= \frac{1}{\sqrt{1-e^{-2x}}}\frac{d}{dx}e^{-x} + \frac{1}{e^{2x}\sqrt{1-e^{-2x}}}\frac{d}{dx}e^x \\ &= -\frac{e^{-x}}{\sqrt{1-e^{-2x}}} + \frac{e^x}{e^{2x}\sqrt{1-e^{-2x}}} \\ &= -\frac{1}{e^x\sqrt{1-e^{-2x}}} + \frac{1}{e^{x}\sqrt{1-e^{-2x}}} \\ \Aboxed{\frac{d}{dx}y&= 0} \end{align} \]