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

2 Answers
Monzur R.
Feb 23, 2017

#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#

marfre
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)] #

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