1. ## differentiation help

Hi
I need to find the derivative of the following
y=(4sqrtx)^lnx
Ummmm, so I'm not really sure how to type that. It is the ("forth root" of x) to the power of (lnx).
I thought that maybe you could change it to
y=x^(.25lnx)
lny=ln.25lnxlnx
lny=ln.25(lnx)^2
(1/y)(y')=f'g+fg'
(1/y)(y')=(0(lnx))+(ln.25)(2(lnx)(1/x))
y'=[(2(ln.25)(lnx))/x]((4sqrtx)^lnx)
Is this assumption incorrect?

2. Originally Posted by sberxa
Hi
I need to find the derivative of the following
y=(4sqrtx)^lnx
Ummmm, so I'm not really sure how to type that. It is the ("forth root" of x) to the power of (lnx).
I thought that maybe you could change it to
y=x^(.25lnx)
lny=ln.25lnxlnx
lny=ln.25(lnx)^2
(1/y)(y')=f'g+fg'
(1/y)(y')=(0(lnx))+(ln.25)(2(lnx)(1/x))
y'=[(2(ln.25)(lnx))/x]((4sqrtx)^lnx)
Is this assumption incorrect?
$y = (\sqrt[4]{x})^{\ln{x}}$

$y = \left(x^{\frac{1}{4}}\right)^{\ln{x}}$

$y = x^{\frac{1}{4}\ln{x}}$

$\ln{y} = \ln{x^{\frac{1}{4}\ln{x}}}$

$\ln{y} = \frac{1}{4}\ln{x}\ln{x}$

$\ln{y} = \frac{1}{4}(\ln{x})^2$

$\ln{y} = \left(\frac{1}{2}\ln{x}\right)^2$

$y = e^{\left(\frac{1}{2}\ln{x}\right)^2}$.

Now use the Chain rule.

You should get

$\frac{dy}{dx} = \frac{\ln{x}e^{(\frac{1}{2}\ln{x})^2}}{2x}$.

3. $\ln{y} = \frac{1}{4}(\ln{x})^2$

$\frac{y'}{y} = \frac{1}{2}(\ln{x}) \frac{1}{x}$

$y' = \frac{y \ln{x}}{2x} = \frac{(\sqrt[4]{x})^{\ln{x}} \ln{x}}{2x}$

4. Originally Posted by Prove It
$y = (\sqrt[4]{x})^{\ln{x}}$

$y = \left(x^{\frac{1}{4}}\right)^{\ln{x}}$

$y = x^{\frac{1}{4}\ln{x}}$

$\ln{y} = \ln{x^{\frac{1}{4}\ln{x}}}$

$\ln{y} = \frac{1}{4}\ln{x}\ln{x}$

$\ln{y} = \frac{1}{4}(\ln{x})^2$

$\ln{y} = \left(\frac{1}{2}\ln{x}\right)^2$

$y = e^{\left(\frac{1}{2}\ln{x}\right)^2}$.

Now use the Chain rule.

You should get

$\frac{dy}{dx} = \frac{\ln{x}e^{(\frac{1}{2}\ln{x})^2}}{2x}$.
Another way to do all that:
$y = (\sqrt[4]{x})^{\ln{x}}$

$y = e^{\ln{(\sqrt[4]{x})^{\ln{x}}}}$

$y = e^{(\frac{1}{2}\ln{x})^2}$