If f(x) = sqrt{x^2 + 1}, g(x) = frac{x + 1}{x | vedclass

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

Question

(C) Step $1$: Find the derivatives of the given functions.$f'(x) = \frac{d}{dx}(x^2 + 1)^{1/2} = \frac{1}{2}(x^2 + 1)^{-1/2} \cdot 2x = \frac{x}{\sqrt

Vedclass · Accepted Answer

$\frac{2}{\sqrt{5}}$

Vedclass · Answer

(C) Step $1$: Find the derivatives of the given functions.$f'(x) = \frac{d}{dx}(x^2 + 1)^{1/2} = \frac{1}{2}(x^2 + 1)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 + 1}}$.$g'(x) = \frac{(x^2 + 1)(1) - (x + 1)(2x)}{(x^2 + 1)^2} = \frac{x^2 + 1 - 2x^2 - 2x}{(x^2 + 1)^2} = \frac{-x^2 - 2x + 1}{(x^2 + 1)^2}$.$h'(x) = \frac{d}{dx}(2x - 3) = 2$.Step $2$: Evaluate the composition $f'(h'(g'(x)))$.Since $h'(x) = 2$ is a constant function, $h'(g'(x)) = 2$ for any $x$.Therefore, $f'(h'(g'(x))) = f'(2)$.Step $3$: Substitute $x = 2$ into $f'(x)$.$f'(2) = \frac{2}{\sqrt{2^2 + 1}} = \frac{2}{\sqrt{5}}$.

If $f(x) = \sqrt{x^2 + 1}$, $g(x) = \frac{x + 1}{x^2 + 1}$, and $h(x) = 2x - 3$, then find the value of $f'(h'(g'(x)))$.

Similar Questions

If $A = \frac{2^x \cot x}{\sqrt{x}}$,then $\frac{dA}{dx} = $

$A$ particle starts from rest and its angular displacement (in radians) is given by $\theta = \frac{t^{2}}{20} + \frac{t}{5}$. If the angular velocity at the end of $t = 4 \ s$ is $k$,then the value of $5k$ is

If $f(x)=3 e^{x^2}$ then $f^{\prime}(x)-2 x f(x)+\frac{1}{3} f(0)-f^{\prime}(0)=$

If $y = f \left( \frac{3x + 4}{5x + 6} \right)$ and $f'(x) = \tan(x^2)$,then $\frac{dy}{dx} = $

If $y = \frac{1}{a - z}$,then $\frac{dz}{dy} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module