Local extrema of y=e^square root of x

To find the local extrema of the function y = e^{\sqrt{x}}, we need to find where the derivative is equal to zero.

Let's start by finding the derivative of y with respect to x:
\frac{dy}{dx} = \frac{d}{dx} (e^{\sqrt{x}})

We can use the chain rule, which states that if u(x) and v(x) are differentiable functions, then the derivative of u(v(x)) with respect to x is u'(v(x)) \cdot v'(x).

Using the chain rule:
\frac{dy}{dx} = e^{\sqrt{x}} \cdot \frac{d}{dx}(\sqrt{x}) = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}}

Now, to find the critical points where the derivative is zero, we set \frac{dy}{dx} = 0:
e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} = 0

This equation has no real solutions for x, which means there are no critical points where the derivative is zero. Therefore, the function y = e^{\sqrt{x}} does not have any local extrema.

\boxed{\text{Answer: There are no local extrema.}}