Euler: Solutions

1. Perhaps you already know that and .

   Then, this shows that D_xcoshx = sinhx.

2. f(x) = x²e^x2

   f'(x) = 2x e^x2 + x² (2x e^x2)

   Don't forget the product rule!

3. h(x) = x^e + e^x + e

   h'(x) = e x^(e-1) + e^x

   e may be a very special number, but it's important to remember that as far a differentiation is concernted, it's just another constant.