How do you differentiate g(x) =sin(3x) (2x+1)^(5/2) using the product rule?

Jan 19, 2018

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

Explanation:

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

$g ' \left(x\right) = f \left(x\right) h ' \left(x\right) + h \left(x\right) f ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{product rule}}$

$\text{differentiate "sin(3x)" and } {\left(2 x + 1\right)}^{\frac{5}{2}}$

$\text{using the "color(blue)"chain rule}$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(g \left(x\right)\right) \times g ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{chain rule}}$

$\frac{d}{\mathrm{dx}} \left(\sin 3 x\right) = \cos \left(3 x\right) \times \frac{d}{\mathrm{dx}} \left(3 x\right) = 3 \cos \left(3 x\right)$

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

$f \left(x\right) = \sin \left(3 x\right) \to f ' \left(x\right) = 3 \cos \left(3 x\right)$

$h \left(x\right) = {\left(2 x + 1\right)}^{\frac{5}{2}} \to h ' \left(x\right) = 5 {\left(2 x + 1\right)}^{\frac{3}{2}}$

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