How do you differentiate #(x^2)arcsin(x^2)#?

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Answer 1 · 2018-04-03T18:09:24

#2xarcsin(x^2)+(2x^3)/sqrt(1-x^4)#

so you have #dot ((uv)) = dot uv+dot vu#

and the derivative of #arcsin(x) = 1/sqrt(1-x^2)#

so if you take #u=x^2# and #v=arcsin(x^2)#
#=> dot u = 2x and dot v = (2x)/sqrt(1-x^4)#

if you apply the values of #u,v,dot u, dot v# to #dot ((uv)) = dot uv+dot vu#

you'll have

the derivative of #x^2arcsin(x^2)# is #2xarcsin(x^2)+(2x^3)/sqrt(1-x^4)#

Answer 2 · 2018-04-03T18:57:42

#f'(x)=2x*arcsin(x^2)+(2x^3)/(sqrt(1-x^4))#

We have:

#f(x)=x^2*arcsin(x^2)#.

#d/dx[f(x)*g(x)]=f'(x)*g(x)+f(x)*g'(x)#

#=>f'(x)=d/dx[x^2]*arcsin(x^2)+x^2*d/dx[arcsin(x^2)]#

#d/dx[f(g(x))]=f'(g(x))*g'(x)#.

#d/dx[x^n]=nx^(n-1)# if #n# is a constant.

#d/dx[arcsin(x)]=1/(sqrt(1-x^2))#

#=>f'(x)=2x*arcsin(x^2)+x^2*1/(sqrt(1-(x^2)^2))*d/dx[x^2]#

#=>f'(x)=2x*arcsin(x^2)+x^2*1/(sqrt(1-x^4))*2x#

#=>f'(x)=2x*arcsin(x^2)+(x^2*2x)/(sqrt(1-x^4))#

#=>f'(x)=2x*arcsin(x^2)+(2x^3)/(sqrt(1-x^4))#