Let f(x) = e^x,g(x) = sin^{-1}x and h(x) = f( | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $f(x) = e^x$ and $g(x) = \sin^{-1}x$. Since $h(x) = f(g(x))$,we have $h(x) = f(\sin^{-1}x) = e^{\sin^{-1}x}$. Now,differentiate $h(x)$ with

Vedclass · Accepted Answer

$1/\sqrt{1 - x^2}$

Vedclass · Answer

(B) Given $f(x) = e^x$ and $g(x) = \sin^{-1}x$. Since $h(x) = f(g(x))$,we have $h(x) = f(\sin^{-1}x) = e^{\sin^{-1}x}$. Now,differentiate $h(x)$ with respect to $x$ using the chain rule: $h'(x) = \frac{d}{dx}(e^{\sin^{-1}x}) = e^{\sin^{-1}x} \cdot \frac{d}{dx}(\sin^{-1}x) = e^{\sin^{-1}x} \cdot \frac{1}{\sqrt{1 - x^2}}$. Therefore,the ratio $h'(x)/h(x)$ is: $\frac{h'(x)}{h(x)} = \frac{e^{\sin^{-1}x} \cdot \frac{1}{\sqrt{1 - x^2}}}{e^{\sin^{-1}x}} = \frac{1}{\sqrt{1 - x^2}}$.

Let $f(x) = e^x$,$g(x) = \sin^{-1}x$ and $h(x) = f(g(x))$,then $h'(x)/h(x) = $

Similar Questions

Find the derivative of the following function: $3 \cot x + 5 \operatorname{cosec} x$

If $y=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots$,then $\frac{dy}{dx}=$ . . . . . . .

$\frac{d}{dx} \sqrt{\frac{1 + \cos 2x}{1 - \cos 2x}} = $

Let $y = \sqrt{p x^3 y}$. If $\frac{dy}{dx} = \frac{3}{2}$ when $y = 1$, then the value of $p$ is equal to

If $y=(1+x)(1+x^2)(1+x^4) \dots (1+x^{2^n})$, then $\left(\frac{dy}{dx}\right)_{x=0}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module