How do you differentiate y=(5x^5+5)(-2x^5-3) using the product rule?

Mar 11, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 100 {x}^{9} - 125 {x}^{4}$

Explanation:

$\text{Given " y=g(x).h(x)" then}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{\frac{\mathrm{dy}}{\mathrm{dx}} = g \left(x\right) h ' \left(x\right) + h \left(x\right) g ' \left(x\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}} \leftarrow \text{ product rule}$

$\text{here } g \left(x\right) = 5 {x}^{5} + 5 \Rightarrow g ' \left(x\right) = 25 {x}^{4}$

$\text{and } h \left(x\right) = - 2 {x}^{5} - 3 \Rightarrow h ' \left(x\right) = - 10 {x}^{4}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \left(5 {x}^{5} + 5\right) \left(- 10 {x}^{4}\right) + \left(- 2 {x}^{5} - 3\right) \left(25 {x}^{4}\right)$

$\textcolor{w h i t e}{\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}} = - 50 {x}^{9} - 50 {x}^{4} - 50 {x}^{9} - 75 {x}^{4}$

$\textcolor{w h i t e}{\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}} = - 100 {x}^{9} - 125 {x}^{4}$