How do you differentiate #f(x)=(t^3)(tan6t)#?

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How do you differentiate #f(x)=(t^3)(tan6t)#?

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Answer 1 · 2015-08-29T16:54:11

If it's the correct question, #0# since differentiating constants give you #0#. If the question is #f(t)=(t^3)(tan 6t)#, then by using the chain rule, you should get #f'(t)=(3t^2)(tan 6t)[1 +(2t sec^2 6t)]#.

Explanation:

#f(t)=(t^3)(tan 6t)#
#f'(t)=(3t^2)(tan 6t) +(t^3)(6 tan 6t sec^2 6t) #
#f'(t)=(3t^2)(tan 6t) +(3t^2 tan 6t)(2t sec^2 6t) #
#f'(t)=(3t^2)(tan 6t)[1 +(2t sec^2 6t)] #

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How do you differentiate #f(x)=(t^3)(tan6t)#?

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