# What is the derivative of f(x)=5x arcsin(x)?

Jul 3, 2018

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

#### Explanation:

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

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

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

$g \left(x\right) = 5 x \Rightarrow g ' \left(x\right) = 5$

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

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

Jul 3, 2018

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

#### Explanation:

Given: $f \left(x\right) = 5 x \arcsin \left(x\right)$

$\textcolor{b l u e}{\text{Building the required relationships}}$

I much prefer the 'old fashioned' notation.

Using $\frac{\mathrm{dy}}{\mathrm{dx}} = u \frac{\mathrm{dv}}{\mathrm{dx}} + v \frac{\mathrm{du}}{\mathrm{dx}}$

Set $y = f \left(x\right) = 5 x \arcsin \left(x\right)$

Set $\textcolor{red}{u = 5 x \implies \frac{\mathrm{du}}{\mathrm{dx}} = 5} \text{ } \ldots \ldots \ldots \ldots \ldots \ldots . . E q u a t i o n \left(1\right)$

Set $v = \arcsin \left(x\right) \implies x = \sin \left(v\right) \implies \frac{\mathrm{dx}}{\mathrm{dv}} = \cos \left(v\right)$

color(white)("dddddddddddddddddddddddddd.d") (dv)/dx=1/cos(v)" ".. Eqn(2)

However: ${\left[\cos \left(v\right)\right]}^{2} + {\left[\sin \left(v\right)\right]}^{2} = 1$

Thus cos(v)=sqrt(1-[sin(v)]^2

From the beginnings of $E q n \left(2\right) \text{ } v = \arcsin \left(x\right)$ thus by substitution:

cos(v)=sqrt(1-[sin(v)]^2) color(white)("dd")=color(white)("dd")sqrt(1-[sin(arcsin(x))]^2

$\textcolor{w h i t e}{\text{dddddddddddddddd.dddd") =color(white)("dd}} \sqrt{1 - {x}^{2}} = \frac{\mathrm{dx}}{\mathrm{dv}}$

Thus: $\textcolor{red}{v = \arcsin \left(x\right) \implies \frac{\mathrm{dv}}{\mathrm{dx}} = \frac{1}{\sqrt{1 - {x}^{2}}}} \text{ } \ldots . . E q n \left({2}_{a}\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Putting it all together}}$

$f ' \left(x\right) = \frac{\mathrm{dy}}{\mathrm{dx}} = \textcolor{w h i t e}{\text{dddd")u(dv)/dxcolor(white)("dddddddd")+color(white)("ddddd}} v \frac{\mathrm{du}}{\mathrm{dx}}$

color(white)("dddddddddd")=[color(white)("d")5x xx1/sqrt(1-x^2)color(white)("d")] +[color(white)(2/2)arcsin(x)xx 5color(white)(2/2)]

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