Derivatives of Inverse Functions

Question

f(x) = 3sin(x)arcsin(x)

f'(x) =

Answer 1

f(x) = 3sin(x) arcsin(x)

[ f(x) ] ' = [ 3sin(x) arcsin(x) ]' =

= [ 3sin(x) ]' arcsin(x) + 3sin(x) [ arcsin(x) ]' =

= 3 cos x arcsin(x) + 3sin(x) [ 1/√(1 - x²)

Answer 2

you will need use derivative product rule

one term will be (sin(x))*d(sin^-1x)/dx

for *d(sin^-1x)/dx, start with sin(sin^-1(x))=x

on left side of equation use chain rule,d(sin(sin^-1(x))/dx=[d(sin(sin^-1(x))/dsin^-1(x)]*[d(sin^-1(x))dx]

on right side dx/dx=1

have cos(sin^-1x)*d(sin^-1(x))dx=1

d(sin^-1(x))dx=1/( cos(sin^-1x))=1/√(1-x²)

f'(x)=3cos(x)*sin^-1(x)+3sin(x)*(1/√(1-x²))