Given that y = exp(2x) * (x^2 +1)^(5/2), what is dy/dx when x is 0?

y = e^2x(x²+1)^5/2

The first step is to calculate dy/dx. We can do this by splitting y into two parts and using the chain rule of differentiation:

y = uv

where u = e^2x and v = (x²+1)^5/2. We now differentiate u and v separately with respect to x.

Here, remember that df(g(x))/dx is equal to df/dg times dg/dx. So,

du/dx = 2e^2x and dv/dx = 5/2 (x²+1)^3/2 2x = 5x(x²+1)^3/2

Using the chain rule,

dy/dx = u dv/dx + v du/dx

= e^2x 5x(x²+1)^3/2 + 2e^2x (x²+1)^5/2.

Now, when x=0, the first term disappears, since it's multipled by x. exp(0) is equal to 1, as is 1^5/2, so the second term reduces to 2 times 1. The answer is therefore 2.