Hi,

To find this derivative you need to use both the product rule and the chain rule. I'm going to illustrate with a similar problem.

Find $D_t \left[ \sin(t^2) cos(1-t) \right].$

You first need to see that this function is a procuct of a sine function and a cosine function so you need to use the product rule.

\[D_t \left[ \sin(t^2) \cos(1-t) \right] = \sin(t^2) D_t \left[\cos(1 - t)\right] + D_t \left[ \sin(t^2) \right] \cos(1-t)\]

To differentiate both $\cos(1 - t)$ and $\sin(t^2)$ you need to use the chain rule.

\[D_t \left[\cos(1 - t)\right] = - \sin(1 - t) D_t[1 - t] = - \sin(1 - t) (-1) = \sin(1 - t)\]

and

\[D_t \left[ \sin(t^2) \right] = \cos(t^2) D_t[t^2] = \cos(t^2) (2t) = 2t \cos(t^2)\]

Thus

\[D_t \left[ \sin(t^2) \cos(1-t) \right] = \sin(t^2) \sin(1 - t) + 2t \cos(t^2) \cos(1-t)\]

Now you try $D_t[\sin t \tan(t^2 + 1)].$

Harley