# How do you find the derivative of 5x arcsin(x)?

Mar 3, 2017

$\frac{5 x}{\sqrt{1 - {x}^{2}}} + 5 \arcsin \left(x\right)$

#### Explanation:

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{\frac{d}{\mathrm{dx}} \left(\arcsin x\right) = \frac{1}{\sqrt{1 - {x}^{2}}}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{let } f \left(x\right) = 5 x \arcsin \left(x\right)$

differentiate f(x) using the $\textcolor{b l u e}{\text{product rule}}$

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

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{f ' \left(x\right) = g \left(x\right) h ' \left(x\right) + h \left(x\right) g ' \left(x\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

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

$\text{and } h \left(x\right) = \arcsin \left(x\right) \Rightarrow h ' \left(x\right) = \frac{1}{\sqrt{1 - {x}^{2}}}$

$\Rightarrow f ' \left(x\right) = 5 x . \frac{1}{\sqrt{1 - {x}^{2}}} + 5 \arcsin \left(x\right)$

$\Rightarrow f ' \left(x\right) = \frac{5 x}{\sqrt{1 - {x}^{2}}} + 5 \arcsin \left(x\right)$