How do you differentiate y= x+((x+sin^2x)^3) ^4?

1 Answer
Jul 12, 2016

y' = 1+12(x + sin^2(x))^11 (1-2sin(x)cos(x))

Explanation:

This problem is solved using the chain rule:
d/dx f(g(x)) = f'(g(x))*g'(x)

y = x + ((x + sin^2(x))^3)^4 = x + (x+sin^2(x))^12

Taking the derivative:
(dy)/dx = d/dx x + d/dx (x+sin^2(x))^12
= 1 + 12(x + sin^2(x))^11*(d/dx(x + sin^2(x)))
= 1 + 12(x + sin^2(x))^11*(d/dx x + d/dx sin^2(x))
= 1 + 12(x + sin^2(x))^11*(1 + 2sin(x)(d/dx sin(x)))
= 1 + 12(x + sin^2(x))^11(1 - 2sin(x)cos(x))