Let Y = 7x sin-1(2x)+^3 square root 1+4x2. Find Y'.

Question

I don't know how to do the square root part.
I have done
y'= (7x)' (Sin^-1(2x)) + (7x)(sin^-1)'+......

y'= 7 sin^-1 + (7x)sq(2x)'/square root (1-(2x))^2 + .....

Answer 1

Does ^3square root mean cube root? I'll assume so. Roots are just fractional exponents, so

y = 7x arcsin(2x)+(1+4x^2)^(1/3)

recall that
d/dx arcsin(u) = 1/√(1-u^2) du/dx, so

Now, using the product rule,

y' = 7 arcsin(2x) + 14x/√(1-4x^2) + (1/3)(1+4x^2)^(-2/3) (8x)
= 7 arcsin(2x) + 14x/√(1-4x^2) + 8x/(3∛(1+4x^2)^2)

Verify this at

http://www.wolframalpha.com/input/?i=derivative+7x+arcsin%282x%29%2B%E2%88%9B%281%2B4x^2%29

If I have misinterpreted something, fix it and work the same steps. You can also fix it at wolframalpha to check your work.