If x=f(theta) and y=g(theta),then frac{d^2 y} | vedclass

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

Question

(C) Given,$x=f(	heta)$ and $y=g(	heta)$.First,find the first derivative $\frac{dy}{dx}$ using the chain rule:$\frac{dx}{d	heta} = f^{\prime}(	heta

Vedclass · Accepted Answer

$\frac{f^{\prime}(	heta) g^{\prime \prime}(	heta)-g^{\prime}(	heta) f^{\prime \prime}(	heta)}{\left(f^{\prime}(	heta)ight)^3}$

Vedclass · Answer

(C) Given,$x=f(	heta)$ and $y=g(	heta)$.First,find the first derivative $\frac{dy}{dx}$ using the chain rule:$\frac{dx}{d	heta} = f^{\prime}(	heta)$ and $\frac{dy}{d	heta} = g^{\prime}(	heta)$.$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{g^{\prime}(	heta)}{f^{\prime}(	heta)}$.Now,differentiate with respect to $x$:$\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{g^{\prime}(	heta)}{f^{\prime}(	heta)} ight) = \frac{d}{d	heta} \left( \frac{g^{\prime}(	heta)}{f^{\prime}(	heta)} ight) \cdot \frac{d	heta}{dx}$.Using the quotient rule:$\frac{d}{d	heta} \left( \frac{g^{\prime}(	heta)}{f^{\prime}(	heta)} ight) = \frac{g^{\prime \prime}(	heta)f^{\prime}(	heta) - g^{\prime}(	heta)f^{\prime \prime}(	heta)}{(f^{\prime}(	heta))^2}$.Since $\frac{d	heta}{dx} = \frac{1}{dx/d	heta} = \frac{1}{f^{\prime}(	heta)}$,we have:$\frac{d^2y}{dx^2} = \frac{g^{\prime \prime}(	heta)f^{\prime}(	heta) - g^{\prime}(	heta)f^{\prime \prime}(	heta)}{(f^{\prime}(	heta))^2} \cdot \frac{1}{f^{\prime}(	heta)} = \frac{f^{\prime}(	heta)g^{\prime \prime}(	heta) - g^{\prime}(	heta)f^{\prime \prime}(	heta)}{(f^{\prime}(	heta))^3}$.

If $x=f(\theta)$ and $y=g(\theta)$,then $\frac{d^2 y}{d x^2}=$

Similar Questions

Derivative of $\sin ^2 x$ with respect to $e^{\cos x}$ is

If $u = \frac{\tan^{-1} x}{\tan^{-1} x + 1}$ and $v = \tan^{-1}(\tan^{-1} x)$,then $\frac{du}{dv} = \dots$

If $x = t + \frac{1}{t}$ and $y = t - \frac{1}{t}$,then $\frac{dy}{dx} = $

If the equation of the normal to the curve $x=\sqrt{t}$,$y=t-\frac{1}{\sqrt{t}}$ at $t=4$ is

If $x = \frac{t^2}{1+t^5}$ and $y = \frac{2t^3}{1+t^5}$ where $t \neq -1$ is a parameter, then find $\frac{dy}{dx}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module