Question

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

Answer 1

`f'(x)=(5x)/(sqrt(1-x^2))+5arcsinx`

Explanation:

`"differentiate using the "color(blue)"product rule"`

`"given "f(x)=g(x)h(x)" then"`

`f'(x)=g(x)h'(x)+h(x)g'(x)larrcolor(blue)"product rule"`

`g(x)=5xrArrg'(x)=5`

`h(x)=arcsinxrArrh'(x)=1/(sqrt(1-x^2))`

`f'(x)=(5x)/(sqrt(1-x^2))+5arcsinx`

Answer 2

`f'(x)=(5x)/sqrt(1-x^2)+5arcsin(x)`

Explanation:

Given: `f(x)=5xarcsin(x)`

`color(blue)("Building the required relationships")`

I much prefer the 'old fashioned' notation.

Using `dy/dx=u(dv)/dx+v(du)/dx`

Set `y=f(x)=5xarcsin(x)`

Set `color(red)(u=5x => (du)/dx=5)" "....................Equation(1)`

Set `v=arcsin(x) => x=sin(v) =>(dx)/(dv)=cos(v)`

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

However: `[cos(v)]^2+[sin(v)]^2=1`

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

From the beginnings of `Eqn(2)" "v=arcsin(x)` thus by substitution:

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

`color(white)("dddddddddddddddd.dddd") =color(white)("dd")sqrt(1-x^2) = (dx)/(dv)`

Thus: `color(red)(v=arcsin(x) =>(dv)/dx=1/sqrt(1-x^2))" ".....Eqn(2_a)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Putting it all together")`

`f'(x)=dy/dx = color(white)("dddd")u(dv)/dxcolor(white)("dddddddd")+color(white)("ddddd")v(du)/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'(x)=(5x)/sqrt(1-x^2)+5arcsin(x)`