The differential coefficient of {x^3} with re | vedclass

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

Question

(B) Let $u = x^3$ and $v = x^2$. We need to find the derivative of $u$ with respect to $v$,which is $\frac{du}{dv}$. Using the chain rule,$\frac{du}{d

Vedclass · Accepted Answer

${{3x} \over 2}$

Vedclass · Answer

(B) Let $u = x^3$ and $v = x^2$. We need to find the derivative of $u$ with respect to $v$,which is $\frac{du}{dv}$. Using the chain rule,$\frac{du}{dv} = \frac{du/dx}{dv/dx}$. First,differentiate $u$ with respect to $x$: $\frac{du}{dx} = \frac{d}{dx}(x^3) = 3x^2$. Next,differentiate $v$ with respect to $x$: $\frac{dv}{dx} = \frac{d}{dx}(x^2) = 2x$. Therefore,$\frac{du}{dv} = \frac{3x^2}{2x} = \frac{3}{2}x$.

The differential coefficient of ${x^3}$ with respect to ${x^2}$ is:

Similar Questions

If $x=\sqrt{2^{\operatorname{cosec}^{-1} t}}$ and $y=\sqrt{2^{\sec ^{-1} t}}, |t| \geq 1$,then $\frac{d y}{d x}=$

If $x = a(\cos \theta + \theta \sin \theta )$ and $y = a(\sin \theta - \theta \cos \theta )$,then $\frac{dy}{dx} = $

If $x = a (t - 1/t)$ and $y = a (t + 1/t)$,where $t$ is the parameter,then $dy/dx = ?$

If $x=a(t+\sin t)$ and $y=a(1-\cos t)$,then find $\frac{d^2 y}{d x^2}$.

If $x=e^\theta(\sin \theta-\cos \theta)$ and $y=e^\theta(\sin \theta+\cos \theta)$,then $\frac{dy}{dx}$ at $\theta=\frac{\pi}{4}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module