Browse Questions

# Differentiate the following w.r.t. $x : e^{sin^{-1} x}$

$\begin{array}{1 1} \large \frac{e^{sin^{-1}x}}{\sqrt(1-x^2)}\\ \large \frac{e^{sin^{-1}x}}{\sqrt(1+x^2)} \\ \large \frac{e^{cos^{-1}x}}{\sqrt(1-x^2)} \\ \large \frac{e^{cos^{-1}x}}{\sqrt(1+x^2)}\end{array}$

Toolbox:
• According to the Chain Rule for differentiation, given two functions $f(x)$ and $g(x)$, and $y=f(g(x)) \rightarrow y' = f'(g(x)).g'(x)$.
• $\; \large \frac{d(sinx)}{dx} $$=\large \frac{1}{\sqrt(1-x^2)} • \; \large \frac{d(e^x)}{dx}$$= e^x$
Given $y = e^{\sin^{-1}x}$
This is of the form $y = f(g(x)$, where $g(x) = \sin^{-1}x$. so, we can apply the chain rule of differentiation.
According to the Chain Rule for differentiation, given two functions $f(x)$ and $g(x)$, and $y=f(g(x)) \rightarrow y' = f'(g(x)).g'(x)$.
$\; \large \frac{d(sinx)}{dx} $$=\large \frac{1}{\sqrt(1-x^2)} If g(x) = \sin^{-1}x, then g'(x) = \large \frac{1}{\sqrt(1-x^2)} \; \large \frac{d(e^x)}{dx}$$= e^x$
Therefore, $y' = d(e^{\sin^{-1}x})\; dx = e^{\sin^{-1}x}\; \large \frac{1}{\sqrt(1-x^2)} = \large \frac{e^{sin^{-1}x}}{\sqrt(1-x^2)}$