Derivative of log and exponential function

Question

How would you find the derivative of:

$$y= e^\sqrt{x} + \ln\sqrt{x}$$

Would I simply use the product rule?

Answer 1

Since $$D_{x}(\sqrt{x}) = \frac{1}{2 \, \sqrt{x}}$$ then \begin{align} D_{x} y(x) &= D_{x} \left[ e^{\sqrt{x}} + \ln(\sqrt{x}) \right] \\ &= e^{\sqrt{x}} \, D_{x}(\sqrt{x}) + \frac{1}{\sqrt{x}} \, D_{x}(\sqrt{x}) \\ &= \frac{1}{2 \, \sqrt{x}} \, \left[ e^{\sqrt{x}} + \frac{1}{\sqrt{x}} \right] \end{align}

Answer 2

Not quite, you want the chain rule. The first one would have the form $e^u \mapsto u'e^u.$ In this case, that's $\frac{1}{2 \sqrt{x}}e^\sqrt{x},$ by just taking the derivative of $\sqrt{x}$. Can you do the other one?

Answer 3

No, but you can use the chain rule:

$$(g(f(x))' = g'(f(x))f'(x).$$

So

$$(e^{\sqrt{x}})' = e^{\sqrt{x}}\frac{1}{2\sqrt{x}}$$

and for the other one, you can either use the chain rule again, or you can use that $\ln(\sqrt{x}) = \frac{1}{2}\ln(x)$.

Answer 4

$$\frac{d}{dx}\left[e^\sqrt x + \ln\sqrt x\right]$$ $$=\frac{d}{dx}\left[e^\sqrt x\right] + \frac{d}{dx}\left[\ln\sqrt x\right]$$ $$=e^\sqrt{x}\frac{d}{dx}\left[\sqrt{x}\right] + \frac{1}{\sqrt{x}}\frac{d}{dx}\left[\sqrt x\right]$$ $$=\frac{d}{dx}\left[\sqrt x\right]\left(e^\sqrt{x}+ \frac{1}{\sqrt{x}}\right)$$ $$=\frac1{2\sqrt{x}}\left(e^\sqrt{x}+ \frac{1}{\sqrt{x}}\right)$$