Evaluate: frac{d}{d x}left[e^{tan ^{-1} x+cot | vedclass

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

Question

(A) Let $y = e^{	an ^{-1} x+\cot ^{-1} x}+\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}$.We know that $	an ^{-1} x + \cot ^{-1} x

Vedclass · Accepted Answer

$\sqrt{a^2-x^2}$

Vedclass · Answer

(A) Let $y = e^{	an ^{-1} x+\cot ^{-1} x}+\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}$.We know that $	an ^{-1} x + \cot ^{-1} x = \frac{\pi}{2}$ for all $x \in \mathbb{R}$.So,the first term becomes $e^{\pi/2}$,which is a constant. The derivative of a constant is $0$.Now,differentiate the remaining terms: $\frac{d}{dx} \left[ \frac{x}{2} \sqrt{a^2-x^2} + \frac{a^2}{2} \sin ^{-1} \frac{x}{a} ight]$.Using the product rule for the first part: $\frac{d}{dx} \left( \frac{x}{2} \sqrt{a^2-x^2} ight) = \frac{1}{2} \sqrt{a^2-x^2} + \frac{x}{2} \cdot \frac{1}{2\sqrt{a^2-x^2}} \cdot (-2x) = \frac{1}{2} \sqrt{a^2-x^2} - \frac{x^2}{2\sqrt{a^2-x^2}} = \frac{a^2-x^2-x^2}{2\sqrt{a^2-x^2}} = \frac{a^2-2x^2}{2\sqrt{a^2-x^2}}$.For the second part: $\frac{d}{dx} \left( \frac{a^2}{2} \sin ^{-1} \frac{x}{a} ight) = \frac{a^2}{2} \cdot \frac{1}{\sqrt{1-(x/a)^2}} \cdot \frac{1}{a} = \frac{a^2}{2} \cdot \frac{1}{\sqrt{(a^2-x^2)/a^2}} \cdot \frac{1}{a} = \frac{a^2}{2} \cdot \frac{a}{\sqrt{a^2-x^2}} \cdot \frac{1}{a} = \frac{a^2}{2\sqrt{a^2-x^2}}$.Adding these results: $\frac{a^2-2x^2}{2\sqrt{a^2-x^2}} + \frac{a^2}{2\sqrt{a^2-x^2}} = \frac{2a^2-2x^2}{2\sqrt{a^2-x^2}} = \frac{2(a^2-x^2)}{2\sqrt{a^2-x^2}} = \sqrt{a^2-x^2}$.

Evaluate: $\frac{d}{d x}\left[e^{\tan ^{-1} x+\cot ^{-1} x}+\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right]$

Similar Questions

If $f(x)=e^{x} g(x)$,$g(0)=4$,and $g^{\prime}(0)=2$,then $f^{\prime}(0)=$

If $f(1)=1$ and $f^{\prime}(1)=3$,then the derivative of $f(f(f(x)))+(f(x))^2$ at $x=1$ is

If the function $f(x)$ is defined by $f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \dots + \frac{x^2}{2} + x + 1$,then $f'(0) = $

Suppose $f(x)$ is differentiable at $x = 1$ and $\mathop {\lim }\limits_{h \to 0} \frac{1}{h}f(1 + h) = 5$,then $f'(1)$ equals

Let $f(x) = x\sqrt{x\sqrt{x\sqrt{x\dots\infty}}}$ where $x > 0$. Then $f'(3)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module