If y = frac{ax + b}{cx + d},then frac{dx}{dy} | vedclass

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

Question

(B) Given $y = \frac{ax + b}{cx + d}$.To find $\frac{dx}{dy}$,we first express $x$ in terms of $y$:$y(cx + d) = ax + b$$cyx + yd = ax + b$$cyx - ax =

Vedclass · Accepted Answer

$\frac{ad - bc}{(a - cy)^2}$

Vedclass · Answer

(B) Given $y = \frac{ax + b}{cx + d}$.To find $\frac{dx}{dy}$,we first express $x$ in terms of $y$:$y(cx + d) = ax + b$$cyx + yd = ax + b$$cyx - ax = b - yd$$x(cy - a) = b - yd$$x = \frac{b - yd}{cy - a} = \frac{yd - b}{a - cy}$.Now,differentiate $x$ with respect to $y$ using the quotient rule $\frac{d}{dy} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}$:$\frac{dx}{dy} = \frac{d(yd - b) \cdot (a - cy) - (yd - b) \cdot d(a - cy)}{dy} \cdot \frac{1}{(a - cy)^2}$$\frac{dx}{dy} = \frac{d(a - cy) - (yd - b)(-c)}{(a - cy)^2}$$\frac{dx}{dy} = \frac{ad - cdy + cdy - bc}{(a - cy)^2}$$\frac{dx}{dy} = \frac{ad - bc}{(a - cy)^2}$.

If $y = \frac{ax + b}{cx + d}$,then $\frac{dx}{dy} = $

Similar Questions

If $y \sec x + \tan x + x^2 y = 0$,then $\frac{dy}{dx} =$

If $\tan^{-1}(x^2 + y^2) = \alpha$,then $\frac{dy}{dx}$ is equal to

If $\log _{10}\left(\frac{x^{3}-y^{3}}{x^{3}+y^{3}}\right)=2$,then $\frac{dx}{dy} = $

If ${x^y} = {y^x},$ then $\frac{dy}{dx} = $

Consider the non-constant differentiable function $f$ of one variable which obeys the relation $\frac{f(x)}{f(y)}=f(x-y)$. If $f^{\prime}(0)=p$ and $f^{\prime}(5)=q$,then $f^{\prime}(-5)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module