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

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