Let f(x)=sin x, g(x)=cos x, h(x)=x^2, then li | vedclass

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

Question

(B) Given $f(x)=\sin x, g(x)=\cos x, h(x)=x^2$. We need to evaluate the limit $L = \lim _{x \rightarrow 1} \frac{f(g(h(x)))-f(g(h(1)))}{x-1}$. This ex

Vedclass · Accepted Answer

$-2 \sin 1 \cos (\cos 1)$

Vedclass · Answer

(B) Given $f(x)=\sin x, g(x)=\cos x, h(x)=x^2$. We need to evaluate the limit $L = \lim _{x \rightarrow 1} \frac{f(g(h(x)))-f(g(h(1)))}{x-1}$. This expression is the definition of the derivative of the composite function $F(x) = f(g(h(x)))$ at $x=1$, i.e.,$F'(1)$. First, find $F(x) = f(g(h(x))) = \sin(\cos(x^2))$. Now, differentiate $F(x)$ with respect to $x$ using the chain rule: $F'(x) = \cos(\cos(x^2)) \cdot \frac{d}{dx}(\cos(x^2)) = \cos(\cos(x^2)) \cdot (-\sin(x^2)) \cdot \frac{d}{dx}(x^2)$. $F'(x) = \cos(\cos(x^2)) \cdot (-\sin(x^2)) \cdot (2x) = -2x \sin(x^2) \cos(\cos(x^2))$. Now, evaluate at $x=1$: $F'(1) = -2(1) \sin(1^2) \cos(\cos(1^2)) = -2 \sin 1 \cos(\cos 1)$. Thus, the limit is $-2 \sin 1 \cos(\cos 1)$. Therefore, option $B$ is correct.

Let $f(x)=\sin x, g(x)=\cos x, h(x)=x^2$, then $\lim _{x \rightarrow 1} \frac{f(g(h(x)))-f(g(h(1)))}{x-1}=$

Similar Questions

Find the derivative of the following function (it is to be understood that $p, q, r$ are fixed non-zero constants): $\frac{p x^{2}+q x+r}{x}$

Let $f(x)$ be a polynomial function such that $f(x)+f^{\prime}(x)+f^{\prime \prime}(x)=x^{5}+64$. Then,the value of $\lim _{x \rightarrow 1} \frac{f(x)}{x-1}$ is:

$\frac{d}{dx} \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)^2 = $

Suppose $f(x)=x(x+3)(x-2)$,where $x \in [-1,4]$. Then,a value of $c$ in $(-1,4)$ satisfying $f^{\prime}(c)=10$ is

The differential coefficient of $\log _{10} x$ with respect to $\log _{x} 10$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module