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

If $f:R \to R$ is a differentiable function and $f(1) = 4$,then the value of $\lim_{x \to 1} \int_4^{f(x)} \frac{2t}{x - 1} dt$ is:

Let $f$ be any function defined on $R$ and let it satisfy the condition $|f(x) - f(y)| \leq |(x - y)^2|$,for all $(x, y) \in R$. If $f(0) = 1$,then:

Let $f(x) = x - \frac{1}{x}$,then $f^{\prime}(-1)$ is

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

If $f(x) = \frac{1+\sec x}{2(\sec x-1)}$ for $0 < x < \frac{\pi}{2}$ and $f^{\prime}(x) = f(x) \cdot g(x)$, then $g(x) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module