Product Rule of Differentiation

Example 1:

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

Solution:

\begin{align*} y&=(2x^3+5)(x-7)\\ \frac{dy}{dx}&=\frac{d}{dx} (2x^3+5)(x-7)\\\\ \because\;\; \frac{d}{dx} uv&=u \frac{dv}{dx}+v \frac{du}{dx}\\\\ \therefore\;\; \frac{dy}{dx}&=(2x^3+5)\frac{d}{dx}(x-7)+(x-7)\frac{d}{dx}(2x^3+5)\\ \frac{dy}{dx}&=(2x^3+5)(\frac{d}{dx}x-\frac{d}{dx}7)+(x-7)(\frac{d}{dx}2x^3+\frac{d}{dx}5)\\ \frac{dy}{dx}&=(2x^3+5)(\frac{d}{dx}x-\frac{d}{dx}7)+(x-7)(2\frac{d}{dx}x^3+\frac{d}{dx}5)\\\\ \because\;\; \frac{d}{dx} x^n&=nx^{n-1} ,\;\;\; \frac{d}{dx} x=1 \;\;\; \textup{and} \;\;\; \frac{d}{dx} c=0\\\\ \frac{dy}{dx}&=(2x^3+5)(1-0)+(x-7)(2(3x^{3-1})+0)\\ \frac{dy}{dx}&=(2x^3+5)(1)+(x-7)(2(3x^2))\\ \frac{dy}{dx}&=2x^3+5+(x-7)(6x^2)\\ \frac{dy}{dx}&=2x^3+5+6x^3-42x^2\\ \frac{dy}{dx}&=8x^3-42x^2+5\\ \end{align*}

Example 2:

$$ y=x^3(x+4)^5,\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y&=x^3(x+4)^5\\ \frac{dy}{dx}&=\frac{d}{dx} x^3(x+4)^5\\\\ \because\;\; \frac{d}{dx} uv&=u \frac{dv}{dx}+v \frac{du}{dx}\\\\ \therefore\;\; \frac{dy}{dx}&=x^3\frac{d}{dx}(x+4)^5+(x+4)^5\frac{d}{dx}x^3\\ \frac{dy}{dx}&=x^3(5(x+4)^4)\frac{d}{dx}(x+4)+(x+4)^5(3x^2)\\ \frac{dy}{dx}&=5x^3(x+4)^4\left(\frac{d}{dx}x+\frac{d}{dx}4\right)+3x^2(x+4)^5\\ \frac{dy}{dx}&=5x^3(x+4)^4(1+0)+3x^2(x+4)^5\\ \frac{dy}{dx}&=5x^3(x+4)^4+3x^2(x+4)^5\\ \frac{dy}{dx}&=x^2(x+4)^4[5x+3(x+4)]\\ \frac{dy}{dx}&=x^2(x+4)^4(5x+3x+12)\\ \frac{dy}{dx}&=x^2(x+4)^4(8x+12)\\ \frac{dy}{dx}&=x^2(8x+12)(x+4)^4\\ \frac{dy}{dx}&=4x^2(2x+3)(x+4)^4\\ \end{align*}