If y = frac{K^{cos^{-1} x}}{1 + K^{cos^{-1} x | vedclass

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

Question

(C) Given that $t = K^{\cos^{-1} x}$.Substituting $t$ into the expression for $y$,we get $y = \frac{t}{1+t}$.To find $\frac{dy}{dt}$,we differentiate

Vedclass · Accepted Answer

$\frac{1}{(1 + K^{\cos^{-1} x})^2}$

Vedclass · Answer

(C) Given that $t = K^{\cos^{-1} x}$.Substituting $t$ into the expression for $y$,we get $y = \frac{t}{1+t}$.To find $\frac{dy}{dt}$,we differentiate $y$ with respect to $t$ using the quotient rule:$\frac{dy}{dt} = \frac{d}{dt} \left( \frac{t}{1+t} \right)$.Using the quotient rule $\frac{d}{dt} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2}$,where $u = t$ and $v = 1+t$:$\frac{dy}{dt} = \frac{(1+t)(1) - t(1)}{(1+t)^2}$.$\frac{dy}{dt} = \frac{1+t-t}{(1+t)^2} = \frac{1}{(1+t)^2}$.Substituting $t = K^{\cos^{-1} x}$ back into the expression,we get:$\frac{dy}{dt} = \frac{1}{(1 + K^{\cos^{-1} x})^2}$.

If $y = \frac{K^{\cos^{-1} x}}{1 + K^{\cos^{-1} x}}$ and $t = K^{\cos^{-1} x}$,then find $\frac{dy}{dt}$.

Similar Questions

If $y = \tan^{-1}\left(\frac{\sin x + \cos x}{\cos x - \sin x}\right)$,then $\frac{dy}{dx}$ is equal to

If $y(x)=\tan ^{-1}\left(\frac{\sqrt{1+a^2 x^2}-1}{a x}\right)$ and $\left(1+a^2 x^2\right) y^{\prime \prime}+g(x) y^{\prime}=0$, then the sum of the roots of the equation $1+a^2 x^2+g(x)=0$ is

If $f'(x) = \sin(\log x)$ and $y = f\left(\frac{2x + 3}{3 - 2x}\right)$,then $\frac{dy}{dx} = $

If $y = \tan^{-1}\left(\frac{4x}{1+5x^2}\right) + \tan^{-1}\left(\frac{3+8x}{8-3x}\right)$,then $\frac{dy}{dx} = $

If $y=\sin ^2\left(\cot ^{-1} \sqrt{\frac{1+x}{1-x}}\right)$,then $\frac{d y}{d x}$ has the value

Vedclass Test Series

Exam Paper Generator

Online Exam Module