Let g(x) be the anti-derivative of f(x). Then | vedclass

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

Question

(C) Given that $g(x)$ is the anti-derivative of $f(x)$, we have $g^{\prime}(x) = f(x)$.We need to find the function $h(x)$ such that $\int h(x) \, dx

Vedclass · Accepted Answer

$\frac{2 f(x) g(x)}{1+(g(x))^2}$

Vedclass · Answer

(C) Given that $g(x)$ is the anti-derivative of $f(x)$, we have $g^{\prime}(x) = f(x)$.We need to find the function $h(x)$ such that $\int h(x) \, dx = \log _e(1+(g(x))^2) + c$.By the definition of an anti-derivative, $h(x) = \frac{d}{dx} [\log _e(1+(g(x))^2) + c]$.Using the chain rule, $\frac{d}{dx} [\log _e(1+(g(x))^2)] = \frac{1}{1+(g(x))^2} \cdot \frac{d}{dx} (1+(g(x))^2)$.$= \frac{1}{1+(g(x))^2} \cdot (2 g(x) \cdot g^{\prime}(x))$.Since $g^{\prime}(x) = f(x)$, we substitute this into the expression:$h(x) = \frac{2 g(x) f(x)}{1+(g(x))^2}$.Thus, the correct option is $C$.

Let $g(x)$ be the anti-derivative of $f(x)$. Then the function for which $\log _e(1+(g(x))^2)+c$ is an anti-derivative is:

Similar Questions

If $f(x) = \frac{x}{1+|x|}$,for $x \in \mathbb{R}$,then $f^{\prime}(0)$ is equal to :

If $r = [2\phi + \cos^2(2\phi + \pi/4)]^{1/2}$,then what is the value of the derivative $dr/d\phi$ at $\phi = \pi/4$?

If $\frac{d}{d x}\left(\frac{1+x^2+x^4}{1+x+x^2}\right)=a x+b$,then $(a, b)=$

Let $f$ and $g$ be differentiable functions satisfying $g'(a) = 2$,$g(a) = b$,and $f \circ g = I$ (identity function). Then $f'(b)$ is equal to

The value of $\frac{d}{{d(\ln x)}}({e^x}{\ln ^2}x)$ at $x=e$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module