Product Rule of Differentiation

Example 1:

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

Solution:

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