derivative of logs and a related rates problem.

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.

Re: derivative of logs and a related rates problem.

Have more faith in yourself. You got 1 and 2 completely right. For problem 3, I suggest logarithmic differentiation. ln(y) = ln(x) ln(ln(x)). 1/y * dy/dx = ln(ln(x))/x + ln(x)/ln(x)x dy/dx = y(ln(ln(x))/x + ln(x)/ln(x)x). Does that make sense to you. I used the same method one would find the derivative of x^x with. I'm gonna have to have a careful look at prob 4.

Re: derivative of logs and a related rates problem.

The reason Wolfram gives the answer of $\displaystyle \displaystyle \begin{align*} -\sin{(x)} \end{align*}$ is because $\displaystyle \displaystyle \begin{align*} e^{\ln{ \left[ \cos{(x)} \right] } } = \cos{(x)} \end{align*}$.

Re: derivative of logs and a related rates problem.

Quote:

Originally Posted by

**ShadowKnight8702** Have more faith in yourself. You got 1 and 2 completely right.

Just to elaborate on this answer, note that $\displaystyle e^{\ln(f(x))}=f(x)$, so OP's answer simplifies to the one wolfram alpha gave.

Re: derivative of logs and a related rates problem.

Quote:

Originally Posted by

**Gusbob** Just to elaborate on this answer, note that $\displaystyle e^{\ln(f(x))}=f(x)$, so OP's answer simplifies to the one wolfram alpha gave.

Better to generalize with x^{logxy} = y

Re: derivative of logs and a related rates problem.

Ah, I'm seeing some things.

1. Glad that I'm right. I wasn't seeing how x would become part of the numerator.

2. Now that you point it out, I do see that $\displaystyle e^{\lncos(x))=cos(x)$ and if I don't change to -tan(x), the cos's cancel and I'm left with -sin(x)

3. I'm seeing where you are going with logarithmic differentiation. I get

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

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

Then you use the product rule.

$\displaystyle \frac{y'}{y}=\frac{(\ln(\ln(x)))}{x}+\frac{\ln(x)} {(\ln(x)x)}$ and then

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

Assuming that is correct, I don't see any simplification for that.

Any pointers on the related rate problem?

Thanks for the help all.

Re: derivative of logs and a related rates problem.

I might have figured out the rate problem, but I'm unsure.

I'm thinking that $\displaystyle \frac{dy}{dt}$ should be -60 mph, because the police car is approaching the intersection.

That would then give

$\displaystyle (1.6)\frac{dx}{dt}=40+72$ or the car is moving at 70 mph.

Anyone agree? Yup, I'm pretty uncertain of myself. :)