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

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Answer 1 · 2016-10-28T14:08:05

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

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

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

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