How do you differentiate $g(x) = (4cosx) / (sin^3x)$ ?

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Answer 1 · 2015-12-05T16:53:01

$-(4sin^2x+12cos^2x)/sin^4x$

Use the [quotient rule.](https://socratic.org/calculus/basic-differentiation-rules/quotient-rule)

$g'(x)=(sin^3xd/dx[4cosx]-4cosxd/dx[sin^3x])/(sin^6x)$

Find each derivative separately.

$d/dx[4cosx]=-4sinx$

$d/dx[sin^3x]=3sin^2xd/dx[sinx]=3sin^2xcosx$

Plug back in.

$g'(x)=(-4sin^4x-12sin^2xcos^2x)/sin^6x$

$g'(x)=(-sin^2x(4sin^2x+12cos^2x))/sin^6x$

$g'(x)=-(4sin^2x+12cos^2x)/sin^4x$

This can also be written as $-4csc^2x-12cot^2xcsc^2x$ .

How do you differentiate g(x) = (4cosx) / (sin^3x) g(x) = (4cosx) / (sin^3x) ?