How do you find the derivative of f(x)=((6x-x^4)/sinx)^6?

1 Answer
Oct 28, 2016

f'(x) = 6 (((6-4x^3)sinx-(6x-x^4)cosx)/(sin^2x))((6x-x^4)/sinx)^5

Explanation:

f(x) = ((6x-x^4)/sinx)^6

By the chain rule, d/dxf(g(x)) =f'(g(x))g'(x) or, dy/dx=dy/(du)(du)/dx.

So f'(x) = 6((6x-x^4)/sinx)^5 d/dx((6x-x^4)/sinx)

For this second derivative we need to use the quotient rule; d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2

So d/dx((6x-x^4)/sinx) = (sinxd/dx(6x-x^4)-(6x-x^4)d/dxsinx)/(sinx)^2
:. d/dx((6x-x^4)/sinx) = (sinx(6-4x^3)-(6x-x^4)cosx)/(sinx)^2

Combining all these results gives us:
So f'(x) = 6{((6x-x^4)/sinx)^5} {(sinx(6-4x^3)-(6x-x^4)cosx)}/(sinx)^2

Hence,
f'(x) = 6 (((6-4x^3)sinx-(6x-x^4)cosx)/(sin^2x))((6x-x^4)/sinx)^5