Question

find the derivative of x-1/2x+7​

Answer 1

Solution:

Given that:

$\tt \longrightarrow y = \dfrac{x - 1}{2x + 7}$

Differentiating both sides with respect to x, we get:

$\tt \longrightarrow \dfrac{dy}{dx} = \dfrac{d}{dx} \bigg( \dfrac{x - 1}{2x + 7} \bigg)$

We know that:

$\bigstar \: \: \underline{ \boxed{ \tt \dfrac{d}{dx} \bigg( \dfrac{u}{v} \bigg) = \dfrac{v \dfrac{du}{dx} -u \dfrac{dv}{dx} }{ {v}^{2} } }}$

Therefore:

$\tt \longrightarrow \dfrac{dy}{dx} =\dfrac{(2x + 7) \dfrac{d}{dx} (x - 1) -(x - 1) \dfrac{d}{dx}(2x + 7)}{(2x + 7)^{2} }$

$\tt \longrightarrow \dfrac{dy}{dx} =\dfrac{(2x + 7) \cdot 1 -(x - 1) \cdot 2}{(2x + 7)^{2} }$

$\tt \longrightarrow \dfrac{dy}{dx} =\dfrac{(2x + 7) -(2x - 2)}{(2x + 7)^{2} }$

$\tt \longrightarrow \dfrac{dy}{dx} =\dfrac{2x + 7 -2x + 2}{(2x + 7)^{2} }$

$\tt \longrightarrow \dfrac{dy}{dx} =\dfrac{9}{(2x + 7)^{2} }$

Which is our required answer.

Learn More:

$\begin{gathered}\boxed{\begin{array}{c|c}\bf f(x)&\bf\dfrac{d}{dx}f(x)\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \sf k&\sf0\\ \\ \sf sin(x)&\sf cos(x)\\ \\ \sf cos(x)&\sf-sin(x)\\ \\ \sf tan(x)&\sf{sec}^{2}(x)\\ \\ \sf cot(x)&\sf-{cosec}^{2}(x)\\ \\ \sf sec(x)&\sf sec(x)tan(x)\\ \\ \sf cosec(x)&\sf-cosec(x)cot(x)\\ \\ \sf\sqrt{x}&\sf\dfrac{1}{2\sqrt{x}}\\ \\ \sf log(x)&\sf\dfrac{1}{x}\\ \\ \sf{e}^{x}&\sf{e}^{x}\end{array}}\\ \end{gathered}$