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

Aug 29, 2015

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