How do you find the derivative of #f(x)=(x^3-8)tan^2(5x-3)#?

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Answer 1 · 2017-01-03T17:41:39

#f'(x)= 3x^2 tan^2 (5x-3) +10(x^3 -8) tan (5x-3) sec^2 (5x-3)#

First apply product rule of differentiation,
#f' (x)= tan^2 (5x-3) d/dx (x^3 -8) + (x^3 -8) d/dx tan^2 (5x-3)#

=#3x^2 tan^2 (5x-3) +(x^3 -8) d/dx tan^2 (5x-3)#

Now apply chain rule to differentiate #tan^2 (5x-3)#. Let 5x-3=t, so that #d/dx tan^2 (5x-3)= d/dt tan^2 t dt/dx#

= #2 tan t sec^2 t d/dx (5x-3)#

= #2 tan (5x-3) sec^2 (5x-3) (5)#

=#10 tan (5x-3) sec^2 (5x-3)#

Then #f'(x)= 3x^2 tan^2 (5x-3) +10(x^3 -8) tan (5x-3) sec^2 (5x-3)#