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

2 Answers
Apr 3, 2018

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

Explanation:

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)

Apr 3, 2018

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

Explanation:

We have:

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

We use the product rule:

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)]

Chain rule:

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

Power rule:

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))