How do you differentiate #f(x)=csc^4x-21cot^2x#?

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Answer 1 · 2017-05-21T04:23:29

#f'(x) = 2 csc^2x cot x (21 - 2csc^2 x)#

Given: #f(x) = csc^4 x - 21 cot^2 x#

Remember that #csc^4 x = (csc x)^4 " and " cot^2 x = (cot x)^2#

Use the Power Rule: #(u^n)' = n u^(n-1) u'#

Let #u = csc x; " " u' = -csc x cot x; " " n = 4#

Let #u = cot x; " " u' = - csc^2 x; " " n = 2#

#f'(x) = 4 csc^3 x (-csc x cot x) - 2(21) cot x (- csc^2 x)#

Simplify:
#f'(x) = -4 csc^4 x cot x + 42 csc^2 x cot x#

#f'(x) = 2 csc^2 cot x (21 - 2 csc^2 x)#