# How do you differentiate g(x) =(2x+1)^(5/2)cos(5x)  using the product rule?

Jun 17, 2017

$g ' \left(x\right) = 5 {\left(2 x + 1\right)}^{\frac{3}{2}} \left(\cos 5 x - \left(2 x + 1\right) \sin 5 x\right)$

#### Explanation:

$\text{differentiate using the "color(blue)"product and chain rules}$

$\text{given " g(x)=f(x).h(x)" then}$

$g ' \left(x\right) = f \left(x\right) h ' \left(x\right) + h \left(x\right) f ' \left(x\right) \leftarrow \text{ product rule}$

$\text{given " f(x)=g(h(x))" then}$

f'(x)=g'(h(x)xxh'(x)larr" chain rule"

$f \left(x\right) = {\left(2 x + 1\right)}^{\frac{5}{2}} \Rightarrow f ' \left(x\right) = \frac{5}{2} {\left(2 x + 1\right)}^{\frac{3}{2}} \times \frac{d}{\mathrm{dx}} \left(2 x + 1\right)$

$\textcolor{w h i t e}{\times \times \times \times \times \times \times \times \times} = 5 {\left(2 x + 1\right)}^{\frac{3}{2}}$

$h \left(x\right) = \cos 5 x \Rightarrow h ' \left(x\right) = - \sin 5 x \times \frac{d}{\mathrm{dx}} \left(5 x\right)$

$\textcolor{w h i t e}{\times \times \times \times \times \times \times \times} = - 5 \sin 5 x$

$\Rightarrow g ' \left(x\right) = {\left(2 x + 1\right)}^{\frac{5}{2}} \left(- 5 \sin 5 x\right) + 5 \cos 5 x {\left(2 x + 1\right)}^{\frac{3}{2}}$

$\textcolor{w h i t e}{\Rightarrow f ' \left(x\right)} = 5 {\left(2 x + 1\right)}^{\frac{3}{2}} \left(\cos 5 x - \left(2 x + 1\right) \sin 5 x\right)$