Question

differentiation of
$y = \frac{2x + 5}{3x - 2}$
​

Answer 1

$\orange{ \boxed{\boxed{\begin{array}{cc}\rm \to \:given \\ \\ \rm \: y = \frac{2x + 5}{3x - 2} \\ \\ \rm \implies \: \frac{dy}{dx} = \frac{d}{dx} ( \frac{2x + 5}{3x - 2} ) \\ \\ \pink{ {\boxed{\begin{array}{cc}\rm \to \: \: we \: know \: that \\ \\ \rm \: \frac{d}{dx} ( \frac{u}{v} ) = \frac{v. \frac{du}{dx} - u. \frac{dv}{dx} }{ {v}^{2} } \end{array}}}} \\ \\ \rm = \frac{(3x - 2). \frac{d}{dx} (2x + 5) - (2x + 5). \frac{d}{dx}(3x - 2) }{ {(3x - 2)}^{2} } \\ \\ \pink{ {\boxed{\begin{array}{cc}\rm \to \:we \: know \: that \\ \\ \hookrightarrow \rm \: \frac{d \: {x}^{n} }{dx} = n {x}^{n - 1} \\ \\ \hookrightarrow \rm \: \frac{d(u \pm \: v)}{dx} = \frac{d \: u}{dx} \pm \frac{d \: v}{dx} \\ \\ \hookrightarrow \rm \: \frac{d \: (constant)}{dx} = 0\end{array}}}} \\ \\ </p><p> \rm = \frac{(3x - 2) \{ \frac{d \: 2x}{dx} + \frac{d \: 5}{d} \} - (2x + 5) \{ \frac{d \: 3x}{dx} - \frac{d \: 2}{dx} \}}{ {(3x - 2)}^{2} } </p><p> \\ \\ \rm = \frac{(3x - 2)(2.1 + 0) - (2x + 5)(3 .1- 0)}{ {(3x - 2)}^{2} } \\ \\ \rm = \frac{2(3x -2) - 3(2x + 5)}{ {(3x - 2)}^{2} } \\ \\ \rm = \frac{2(3x - 2)}{ {(3x - 2)}^{2} } - \frac{3(2x + 5)}{ {( 3x - 2)}^{2} } \\ \\ \rm = \frac{2}{3x - 2} - \frac{3(2x + 5)}{ {(3x - 2)}^{2} } \\ \\ \sf = answer\end{array}}}}</p><p>$