



 
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 $D_t \left[ \sin(t^2) cos(1t) \right].$ 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. \[D_t \left[ \sin(t^2) \cos(1t) \right] = \sin(t^2) D_t \left[\cos(1  t)\right] + D_t \left[ \sin(t^2) \right] \cos(1t)\] 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 \[D_t \left[ \sin(t^2) \cos(1t) \right] = \sin(t^2) \sin(1  t) + 2t \cos(t^2) \cos(1t)\] Now you try $D_t[\sin t \tan(t^2 + 1)].$ Harley  


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. 