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

#"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#

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)#