If x = frac{9t^2}{1+t^4} and y = frac{16t^2}{ | vedclass

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

Question

(D) Given $x = \frac{9t^2}{1+t^4}$ and $y = \frac{16t^2}{1-t^4}$.First,differentiate $y$ with respect to $t$ using the quotient rule:$\frac{dy}{dt} =

Vedclass · Accepted Answer

$\frac{16}{9}\left(\frac{1+t^4}{1-t^4}\right)^3$

Vedclass · Answer

(D) Given $x = \frac{9t^2}{1+t^4}$ and $y = \frac{16t^2}{1-t^4}$.First,differentiate $y$ with respect to $t$ using the quotient rule:$\frac{dy}{dt} = \frac{(1-t^4)(32t) - (16t^2)(-4t^3)}{(1-t^4)^2} = \frac{32t - 32t^5 + 64t^5}{(1-t^4)^2} = \frac{32t(1+t^4)}{(1-t^4)^2}$.Next,differentiate $x$ with respect to $t$ using the quotient rule:$\frac{dx}{dt} = \frac{(1+t^4)(18t) - (9t^2)(4t^3)}{(1+t^4)^2} = \frac{18t + 18t^5 - 36t^5}{(1+t^4)^2} = \frac{18t(1-t^4)}{(1+t^4)^2}$.Now,find $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$:$\frac{dy}{dx} = \frac{32t(1+t^4)}{(1-t^4)^2} 	imes \frac{(1+t^4)^2}{18t(1-t^4)} = \frac{32(1+t^4)^3}{18(1-t^4)^3} = \frac{16}{9}\left(\frac{1+t^4}{1-t^4}ight)^3$.

If $x = \frac{9t^2}{1+t^4}$ and $y = \frac{16t^2}{1-t^4}$,then $\frac{dy}{dx} = $

Similar Questions

If $x = t + \frac{1}{t}$ and $y = t - \frac{1}{t}$,then $\frac{d^2y}{dx^2}$ is equal to

If $x=\log _e\left(\frac{\cos \frac{y}{2}-\sin \frac{y}{2}}{\cos \frac{y}{2}+\sin \frac{y}{2}}\right)$ and $\tan \frac{y}{2}=\sqrt{\frac{1-t}{1+t}}$,then the value of $\left(\frac{dy}{dx}\right)_{t=\frac{1}{2}}$ is

If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{dy}{dx}$ for $x=a(\theta-\sin \theta)$ and $y=a(1+\cos \theta)$.

The rate of change of $\sqrt{x^2+16}$ with respect to $\frac{x}{x-1}$ at $x=5$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module