frac{d}{d x} left{ (1+x^2) tan^{-1}(x) right} | vedclass

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

Question

(C) To find the derivative of the product $(1+x^2) 	an^{-1}(x)$,we use the product rule: $\frac{d}{dx}(u \cdot v) = u \cdot \frac{dv}{dx} + v \cdot \

Vedclass · Accepted Answer

$2 x 	an^{-1}(x) + 1$

Vedclass · Answer

(C) To find the derivative of the product $(1+x^2) 	an^{-1}(x)$,we use the product rule: $\frac{d}{dx}(u \cdot v) = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx}$.Let $u = (1+x^2)$ and $v = 	an^{-1}(x)$.Then,$\frac{du}{dx} = 2x$ and $\frac{dv}{dx} = \frac{1}{1+x^2}$.Applying the product rule:$\frac{d}{dx} \left\{ (1+x^2) 	an^{-1}(x) ight\} = (1+x^2) \cdot \frac{d}{dx}(	an^{-1}(x)) + 	an^{-1}(x) \cdot \frac{d}{dx}(1+x^2)$$= (1+x^2) \cdot \frac{1}{1+x^2} + 	an^{-1}(x) \cdot (2x)$$= 1 + 2x 	an^{-1}(x)$.Thus,the correct option is $C$.

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

Similar Questions

The differential of $f(x) = \log_{e}(1 + e^{10x}) - \tan^{-1}(e^{5x})$ at $x = 0$ and for $dx = 0.2$ is

Find the derivative of $x^{5}(3-6x^{-9}).$

Differentiate the following with respect to $x:$ $\cos^{-1}(e^x)$

If $f(x) = x^3 + e^{x/2}$ and $g(x) = f^{-1}(x)$,then the value of $g'(1)$ is:

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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module