Is this right?

h(x) = 3e^sin(x + 2)

h'(x) = 3e^sin(x + 2) X cos(x+2)(1)

Is this right?

h(x) = 3e^sin(x + 2)

h'(x) = 3e^sin(x + 2) X cos(x+2)(1)


Yep, that's it. Very good. You'd get a greenie if I could give you one. :)

Except, try using LaTex if you plan on frequenting this math site on a regular basis.

Like so: 3cos(x+2)\cdot e^{sin(x+2)}

To see the code I typed to make it display that way, just click on 'quote user'.