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 $t$ is a parameter and $x=t+\frac{1}{t}, y=t-\frac{1}{t}$, then $\frac{d^2 y}{d x^2}=$

If $\tan u=\sqrt{\frac{1-x}{1+x}}$ and $\cos v=4 x^{3}-3 x$,then $\frac{d u}{d v}=$

What is the slope of the tangent to the curve $x = t^2 + 3t - 8$ and $y = 2t^2 - 2t - 5$ at the point $(2, -1)$?

The derivative of $\sin^2 x$ with respect to $\cos^2 x$ is

The equation of the normal to the curve $x=\theta+\sin \theta, y=1+\cos \theta$ at $\theta=\frac{\pi}{2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module