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

2 Answers
Feb 23, 2017

3sec^2x - 5csc^2x3sec2x5csc2x

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

Feb 23, 2017

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)]