y' of a hyperbolic inverse function?

Question

y = x^2 * sinh^-1(5x)
y' = ?
I come up with (2x)(sinh^-1(5x) + (5x^2) / (1+(5x)^2)^(1/2)

But i Keep getting a wrong input can someone please help?
I may need to simplify but not shure how

Answer 1

Using the product rule:

sinh^-1(5x) ( x^2)' + (x^2) (sinh^-1(5x)' <- notice the primes

using the table of integrals, you find that

d/dx sinh^-1(u) = 1/( (1 - u ^2)^(1/2) )

which when you do it for your equation:
d/dx Sinh^-1(5x)

5/ (1- (5x)^2)^- (1/2)

putting this altogether, you get what you have.

Answer 2

you got it right

y = x^2*sinh^-1(5x)

let u = x^2 : du = 2x

v = sinh^-1(5x) : dv = 1/sqrt[1+(5x^)2] = 1/sqrt(1+25x^2)*(5)

d(uv) = udv + vdu

y' = x^2[5/sqrt(1+25x^2)] +sinh^-1(5x)*(2x)

y' = 5x^2/sqrt(1+25x^2) + 2x sinh^-1(5x)

Answer 3

y' = 5x^2/sqrt(1+25x^2) +2xsinh^-1(5x)

Your answer is correct