Question

$\large\underline{\purple{\mathfrak{Differentiation}}} :$

$\textsf{ Use \red{chain\; rule} to find \frac{dy}{dx}} ,$
$\\\bf if \: y = \left(\dfrac{2x-1}{2x+1}\right)^2$

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Answer 1

$\sf\underline{\red{Proper\: Question}}:$

• Use chain rule to find dy/dx,$\bf if \: y = \left(\dfrac{2x-1}{2x+1}\right)^2$

$\sf\underline{\red{Solution}}:$

⇨ Let y = u² , $\sf where\: u = \dfrac{2x-1}{2x+1}$

$\bf Then \:\:\:\:\:\:\:\: \frac{dy}{dx} = 2u$

and

$\longmapsto \dfrac{du}{dx} = \dfrac{(2x+1).\frac{d}{dx}(2x-1)-(2x-1).\frac{d}{dx}(2x+1)}{(2x+1)^2} \\\\\\\implies \dfrac{du}{dx} = \dfrac{(2x+1).(2)-(2x-1).(2)}{(2x+1)^2} \\\\\\\implies \dfrac{du}{dx} = \dfrac{\cancel{4x}+2\: \cancel{- 4x}+2}{(2x+1)^2} \\\\\\\implies\blue{ \dfrac{du}{dx}} =\blue{\dfrac{4}{(2x+1)^2}}$

$\textbf{By chain rule}, \dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dx}$

$\therefore \dfrac{dy}{dx} = 2u.\dfrac{4}{(2x+1)^2} \\\\\\\mapsto \sf Put\:the\:value\:of\:u\\\\\implies \dfrac{dy}{dx} = 2.\dfrac{2x-1}{2x+1}.\dfrac{4}{(2x+1)^2} \\\\\\\implies \dfrac{dy}{dx} = \dfrac{8(2x-1)}{(2x+3)^3} , x \neq- \frac{1}{2}$

$\bf \underline{Hence} ,$

$\\\:\:\:\:\:\:\dfrac{dy}{dx} = \frac{8(2x-1)}{(2x+3)^3} , x \neq- \dfrac{1}{2}$