Find the derivative: frac{d}{dx} left( frac{e | vedclass

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

Question

(B) To find the derivative of $\frac{e^x}{1 + x^2}$,we use the quotient rule: $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \fr

Vedclass · Accepted Answer

$\frac{e^x(1 - x)^2}{(1 + x^2)^2}$

Vedclass · Answer

(B) To find the derivative of $\frac{e^x}{1 + x^2}$,we use the quotient rule: $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$.Let $u = e^x$ and $v = 1 + x^2$.Then $\frac{du}{dx} = e^x$ and $\frac{dv}{dx} = 2x$.Applying the quotient rule:$\frac{d}{dx} \left( \frac{e^x}{1 + x^2} \right) = \frac{(1 + x^2)e^x - e^x(2x)}{(1 + x^2)^2}$.Factor out $e^x$ from the numerator:$= \frac{e^x(1 + x^2 - 2x)}{(1 + x^2)^2}$.Recognizing that $1 + x^2 - 2x = (1 - x)^2$:$= \frac{e^x(1 - x)^2}{(1 + x^2)^2}$.

Find the derivative: $\frac{d}{dx} \left( \frac{e^x}{1 + x^2} \right)$

Similar Questions

If $y = f\left( \frac{2x - 1}{x^2 + 1} \right)$ and $f'(x) = \sin(x^2),$ then $\frac{dy}{dx} = $

If $y = b \cos \log \left( \frac{x}{n} \right)^n$,then $\frac{dy}{dx} = $

$\frac{d}{dx}\left[ \frac{2}{\pi }\sin {x^\circ} \right] = $

Suppose a differentiable function $f(x)$ satisfies the identity $f(x+y) = f(x) + f(y) + xy^2 + x^2y$ for all real $x$ and $y$. If $\lim_{x \rightarrow 0} \frac{f(x)}{x} = 1$,then $f'(3)$ is equal to:

If $y = \frac{a + b{x^{3/2}}}{{x^{5/4}}}$ and $y' = 0$ at $x = 5$,then the ratio $a:b$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module