Question

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

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

$\green{\textbf{ Spammers stay away}}$

Answer 1

Step-by-step explanation:

We have,

$y = \bigg( \frac{2x - 1}{2x + 1} \bigg)^{2}$

$\implies\frac{dy}{dx} = 2\bigg( \frac{2x - 1}{2x + 1} \bigg). \frac{d}{dx} \bigg( \frac{2x - 1}{2x + 1} \bigg)$

$\implies\frac{dy}{dx} = 2\bigg( \frac{2x - 1}{2x + 1} \bigg). \frac{(2x + 1). \frac{d}{dx}(2x - 1) - (2x - 1). \frac{d}{dx}(2x + 1) }{ {(2x + 1)}^{2} }$

$\implies\frac{dy}{dx} = 2\bigg( \frac{2x - 1}{2x + 1} \bigg). \frac{2(2x + 1) - 2(2x - 1). }{ {(2x + 1)}^{2} }$

$\implies\frac{dy}{dx} = 2\bigg( \frac{2x - 1}{2x + 1} \bigg). \frac{4x + 2 - 4x + 2 }{ {(2x + 1)}^{2} }$

$\implies\frac{dy}{dx} = 2\bigg( \frac{2x - 1}{2x + 1} \bigg). \frac{4}{ {(2x + 1)}^{2} }$

$\implies\frac{dy}{dx} = 8 \bigg \{ \frac{2x - 1}{(2x + 1) ^{3} } \bigg \}$

Answer 2

Answer:

$Step-by-step explanation:</p><p></p><p>We have,</p><p></p><p>\begin{gathered}y = \bigg( \frac{2x - 1}{2x + 1} \bigg)^{2} \\ \end{gathered}y=(2x+12x−1)2</p><p></p><p>\begin{gathered} \implies\frac{dy}{dx} = 2\bigg( \frac{2x - 1}{2x + 1} \bigg). \frac{d}{dx} \bigg( \frac{2x - 1}{2x + 1} \bigg)\\ \end{gathered}⟹dxdy=2(2x+12x−1).dxd(2x+12x−1)</p><p></p><p>\begin{gathered} \implies\frac{dy}{dx} = 2\bigg( \frac{2x - 1}{2x + 1} \bigg). \frac{(2x + 1). \frac{d}{dx}(2x - 1) - (2x - 1). \frac{d}{dx}(2x + 1) }{ {(2x + 1)}^{2} } \\ \end{gathered}⟹dxdy=2(2x+12x−1).(2x+1)2(2x+1).dxd(2x−1)−(2x−1).dxd(2x+1)</p><p></p><p>\begin{gathered} \implies\frac{dy}{dx} = 2\bigg( \frac{2x - 1}{2x + 1} \bigg). \frac{2(2x + 1) - 2(2x - 1). }{ {(2x + 1)}^{2} } \\ \end{gathered}⟹dxdy=2(2x+12x−1).(2x+1)22(2x+1)−2(2x−1).</p><p></p><p>\begin{gathered} \implies\frac{dy}{dx} = 2\bigg( \frac{2x - 1}{2x + 1} \bigg). \frac{4x + 2 - 4x + 2 }{ {(2x + 1)}^{2} } \\ \end{gathered}⟹dxdy=2(2x+12x−1).(2x+1)24x+2−4x+2</p><p></p><p>\begin{gathered} \implies\frac{dy}{dx} = 2\bigg( \frac{2x - 1}{2x + 1} \bigg). \frac{4}{ {(2x + 1)}^{2} } \\ \end{gathered}⟹dxdy=2(2x+12x−1).(2x+1)24</p><p></p><p>\begin{gathered} \implies\frac{dy}{dx} = 8 \bigg \{ \frac{2x - 1}{(2x + 1) ^{3} } \bigg \} \\ \end{gathered}⟹dxdy=8{(2x+1)32x−1}</p><p></p><p>$