I'm struggling a bit with log derivatives.

Problem #1.

y= $\displaystyle \sqrt[4]{\ln(3x)}$

I get :

y' =$\displaystyle \frac{1}{4(\ln(3x))^{3/4}}(\frac{1}{3x})(3)$ which simplifies to $\displaystyle \frac{1}{4x(\ln(3x))^{3/4}}$

Mathway.com has $\displaystyle \frac{x}{4(\ln(3x))^{3/4}}$ and Wolframalpha.com agrees with me. So who is correct?

Problem # 2

y= $\displaystyle e^\ln(cos(x))$

The answer I get is y'= $\displaystyle e^{\ln(cos(x))}(\frac{1}{cos(x)})(-sin(x))$ which simplifies to $\displaystyle e^{\ln(cos(x))}(-tan(x))$

Wolfram's says the answer is simply -sin(x). Am I close?

Problem # 3.

y= $\displaystyle (\ln(x))^{ln(x)}$

I get y'= $\displaystyle \ln(x)(\frac{1}{x})$

I'm getting confused with this one.

Problem # 4

The related rate problem I have is:

A police cruiser approaching a intersection from the north, and is chasing a speeding car that is moving east. When the cruiser is .6 miles north of the intersection, and the car is .8 miles east of the intersection, the distance between the cruiser and the car is increasing at 20mph. If the cruiser is moving at 60 mpg at the instant of measurement, what is the speed of the car?

I know I'm doing something wrong here.

I set y= .6, x= .8, z=1 (hypotenuse), $\displaystyle \frac{dy}{dt}$ = 60mph, $\displaystyle \frac{dz}{dt}$= 20mph

The equation I am using is $\displaystyle x^2 + y^2 = z^2$. I take the derivative of both sides and get $\displaystyle 2x(\frac{dx}{dt})+2y(\frac{dy}{dt})=2z(\frac{dz}{d t})$ Substitute 60mph for $\displaystyle \frac{dy}{dt}$. and 20mph for $\displaystyle \frac{dz}{dt}$ and solve for $\displaystyle \frac{dx}{dt}$ and I get $\displaystyle (2(.6)(60))+2(.8)(\frac{dx}{dt})=2(1)(20)$, then $\displaystyle 1.6(\frac{dx}{dt})=40-72$, so $\displaystyle \frac{dx}{dt}= -20mph$

I'm pretty sure the car being chased isn't going -20 mph. Can someone give me the equation I should be using? I'm pretty lost with related rate problems.

I appreciate any help.