Question

How do you find the derivative of `f(x)=sec(3x)csc(5x)`?

Answer 1

`3sec^2x - 5csc^2x`

Explanation:

We need to use the product rule to differentiate this function:

For #f(x) = g(x)h(x), f'(x) = h(x)g'(x) + g(x)h'(x)

For the derivatives of the reciprocal trig functions, see
https://www.khanacademy.org/math/ap-calculus-ab/differentiating-common-functions-ab/trigonometric-functions-differentiation-ab/v/derivatives-of-secx-and-cscx

`f'(x) = 3cscxtanxsecx - 5secxcotxcscx =`

`3secxsecx -5cscxcscx =`

`3sec^2x - 5csc^2x`

Answer 2

`csc(5x) sec(3x) [3tan(3x) - 5 cot(5x)]`

Explanation:

Use need to use the product rule: `(uv)' = uv' + vu'`

And the chain rule:
`(sec u)' = (sec u tan u)u'`

`(csc u)' = (-csc u cot u)u'`

Applying the product rule for the function: `f(x) = sec(3x)csc(5x)`
let `u = sec(3x)` and `v = csc(5x)`

For the chain rule on `sec(3x)` let `u = 3x`, so `u' = 3`

For the chain rule on `csc(5x)` let `u = 5x`, so `u' = 5`

`f'(x) = sec(3x)[-csc(5x)cot(5x)] (5) + csc(5x)[sec(3x)tan(3x)] (3)`

Simplify: `f'(x) = -5sec(3x) csc(5x) cot(5x) + 3csc(5x) sec(3x) tan(3x)`

Rearrange:
`f(x)' = 3 csc(5x) sec(3x) tan(3x) - 5 csc(5x) sec(3x) cot(5x)`

Factor: `f(x)' = csc(5x) sec(3x) [3tan(3x) - 5 cot(5x)]`