frac{d}{dx} { e^x log(1 + x^2) } = | vedclass

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

Question

(A) To find the derivative of the product $e^x \log(1 + x^2)$,we use the product rule: $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$.Let $u =

Vedclass · Accepted Answer

$e^x \left[ \log(1 + x^2) + \frac{2x}{1 + x^2} \right]$

Vedclass · Answer

(A) To find the derivative of the product $e^x \log(1 + x^2)$,we use the product rule: $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$.Let $u = e^x$ and $v = \log(1 + x^2)$.Then $\frac{du}{dx} = e^x$ and $\frac{dv}{dx} = \frac{1}{1 + x^2} \cdot \frac{d}{dx}(1 + x^2) = \frac{2x}{1 + x^2}$.Applying the product rule:$\frac{d}{dx} \{ e^x \log(1 + x^2) \} = e^x \cdot \frac{d}{dx}(\log(1 + x^2)) + \log(1 + x^2) \cdot \frac{d}{dx}(e^x)$$= e^x \left( \frac{2x}{1 + x^2} \right) + \log(1 + x^2) \cdot e^x$$= e^x \left[ \log(1 + x^2) + \frac{2x}{1 + x^2} \right]$.

$\frac{d}{dx} \{ e^x \log(1 + x^2) \} = $

Similar Questions

Differentiate the following with respect to $x$: $\sin \left(\tan ^{-1} e^{-x}\right)$

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

Differentiate $\sin \left(\cos \left(x^{2}\right)\right)$ with respect to $x.$

If $y = (1 + x^2) \tan^{-1} x - x$,then $\frac{dy}{dx}$ is

If $y = \sec(\tan^{-1} x)$,then $\frac{dy}{dx}$ at $x = 1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module