How do you differentiate  f(x)=e^sqrt(lnsqrtx) using the chain rule.?

Mar 3, 2018

$f ' \left(x\right) = \frac{{e}^{\sqrt{\ln \sqrt{x}}}}{4 \sqrt{\ln \sqrt{x}}}$

Explanation:

We are given: $f \left(x\right) = {e}^{\sqrt{\ln \sqrt{x}}}$

Chain rule is: $\frac{\mathrm{dg} \left(h \left(x\right)\right)}{\mathrm{dx}} = \frac{\mathrm{dg} \left(h \left(x\right)\right)}{\mathrm{dh} \left(x\right)} \cdot \frac{\mathrm{dh} \left(x\right)}{\mathrm{dx}}$

If we take our equation and try to match forms, we see:
$g \left(h \left(x\right)\right) = f \left(x\right) = {e}^{\sqrt{\ln \sqrt{x}}}$
$h \left(x\right) = \sqrt{\ln \sqrt{x}}$

Now we can use the chain rule:
f'(x) = (d(e^sqrt(lnsqrtx)))/(d(sqrt(lnsqrt(x))))*(d(sqrt(lnsqrtx)))/(dx

For convenience, let $\setminus \tau \setminus \equiv \sqrt{\ln \sqrt{x}}$ .

Let's do the first term:
$\frac{d \left({e}^{\setminus \tau}\right)}{d \setminus \tau} = {e}^{\setminus \tau} = {e}^{\sqrt{\ln \sqrt{x}}}$

Now the second term:
$\frac{d \setminus \tau}{\mathrm{dx}} = \frac{d \left(\sqrt{\ln \sqrt{x}}\right)}{\mathrm{dx}} = \frac{1}{4 \sqrt{\ln \sqrt{x}}}$

Hence:
$f ' \left(x\right) = \frac{{e}^{\sqrt{\ln \sqrt{x}}}}{4 \sqrt{\ln \sqrt{x}}}$