Differentiate the following with respect to x | vedclass

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

Question

(A) Let $y = e^{x^{3}}$.To differentiate $y$ with respect to $x$,we use the chain rule.$\frac{dy}{dx} = \frac{d}{dx}(e^{x^{3}})$.Applying the chain ru

Vedclass · Accepted Answer

$3x^{2}e^{x^{3}}$

Vedclass · Answer

(A) Let $y = e^{x^{3}}$.To differentiate $y$ with respect to $x$,we use the chain rule.$\frac{dy}{dx} = \frac{d}{dx}(e^{x^{3}})$.Applying the chain rule,we treat $x^{3}$ as a function $u$,so $\frac{d}{dx}(e^{u}) = e^{u} \cdot \frac{du}{dx}$.$\frac{dy}{dx} = e^{x^{3}} \cdot \frac{d}{dx}(x^{3})$.Since $\frac{d}{dx}(x^{3}) = 3x^{2}$,we get:$\frac{dy}{dx} = e^{x^{3}} \cdot 3x^{2} = 3x^{2}e^{x^{3}}$.

Differentiate the following with respect to $x$: $e^{x^{3}}$

Similar Questions

If $y = \tan^{-1} \sqrt{\frac{1 + \sin 2x}{1 - \sin 2x}}$, then at $x = \frac{\pi}{6}$, the value of $\frac{dy}{dx}$ is

Let $f(x + y) = f(x)f(y)$ and $f(x) = 1 + \sin(3x)g(x)$,where $g(x)$ is continuous. Then $f'(x)$ is:

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:

Find the derivative of $f$ given by $f(x) = \tan^{-1} x$,assuming it exists.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module