Quotient Rule of Differentiation

Example 1:

$$ y=\frac{(2x^3+5)}{(x-7)},\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y&=\frac{(2x^3+5)}{(x-7)}\\ \frac{dy}{dx}&=\frac{d}{dx} \frac{(2x^3+5)}{(x-7)}\\\\ \because\;\; \frac{d}{dx} \frac{u}{v}&= \frac{v \frac{du}{dx}-u \frac{dv}{dx}}{v^2}\\\\ \therefore\;\; \frac{dy}{dx}&=\frac{(x-7)\frac{d}{dx}(2x^3+5)-(2x^3+5)\frac{d}{dx}(x-7)}{(x-7)^2}\\ \frac{dy}{dx}&=\frac{(x-7)(\frac{d}{dx}2x^3+\frac{d}{dx}5)-(2x^3+5)(\frac{d}{dx}x-\frac{d}{dx}7)}{(x-7)^2}\\ \frac{dy}{dx}&=\frac{(x-7)(2\frac{d}{dx}x^3+\frac{d}{dx}5)-(2x^3+5)(\frac{d}{dx}x-\frac{d}{dx}7)}{(x-7)^2}\\\\ \because\;\; \frac{d}{dx} x^n&=nx^{n-1} ,\;\;\; \frac{d}{dx} x=1 \;\;\; \textup{and} \;\;\; \frac{d}{dx} c=0\\\\ \frac{dy}{dx}&=\frac{(x-7)(2(3x^{3-1})+0)-(2x^3+5)(1-0)}{(x-7)^2}\\ \frac{dy}{dx}&=\frac{(x-7)(2(3x^2))-(2x^3+5)(1)}{(x-7)^2}\\ \frac{dy}{dx}&=\frac{(x-7)(6x^2)-(2x^3+5)}{(x-7)^2}\\ \frac{dy}{dx}&=\frac{6x^3-42x^2-2x^3-5}{(x-7)^2}\\ \frac{dy}{dx}&=\frac{4x^3-42x^2-5}{(x-7)^2}\\ \end{align*}

Example 2:

$$ y=\frac{4x}{x+5},\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y&=\frac{4x}{x+5}\\ \frac{dy}{dx}&=\frac{d}{dx} \frac{4x}{x+5}\\\\ \because\;\; \frac{d}{dx} \frac{u}{v}&= \frac{v \frac{du}{dx}-u \frac{dv}{dx}}{v^2}\\\\ \therefore\;\; \frac{dy}{dx}&=\frac{(x+5)\frac{d}{dx}(4x)-(4x)\frac{d}{dx}(x+5)}{(x+5)^2}\\ \frac{dy}{dx}&=\frac{(x+5)(4\frac{d}{dx}x)-(4x)(\frac{d}{dx}x+\frac{d}{dx}5)}{(x+5)^2}\\\\ \because\;\; \frac{d}{dx} x&=1 \;\;\; \textup{and} \;\;\; \frac{d}{dx} c=0\\\\ \frac{dy}{dx}&=\frac{(x+5)(4(1))-(4x)(1+0)}{(x+5)^2}\\ \frac{dy}{dx}&=\frac{(x+5)(4)-(4x)(1)}{(x+5)^2}\\ \frac{dy}{dx}&=\frac{4x+20-4x}{(x+5)^2}\\ \frac{dy}{dx}&=\frac{20}{(x+5)^2}\\ \end{align*}