How to find the derivative of this?

[ab94]

Homework Statement

Find dy/dx of this:
Assume p>0, y=(p^(x^p))(x^p)

Homework Equations

1. d/dx f(x)g(x)h(x) = f'(x)g(x)h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x)

The Attempt at a Solution

ln(y)= p(ln(p^x))x^p

1/y y'=p'(ln(p^x))x^p + p(logp)x^p + p(ln(p^x))px^(p-1) --use "revlevent" equation

1/y y'= 0 + plogp(x^p) + p(ln(p^x))px^(p-1) -- simplify derivatives

y'= y(plogp(x^p) + p(ln(p^x))px^(p-1)) -- multiply both sides by y

y'= (p^(x^p))(x^p)(plogp(x^p) + p(ln(p^x))px^(p-1)) -- sub in y from beginning




Can someone point out where my error is, and how to fix it?

Thanks

 

[SammyS]

Homework Statement

Find dy/dx of this:
Assume p>0, y=(p^(x^p))(x^p)

Homework Equations

1. d/dx f(x)g(x)h(x) = f'(x)g(x)h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x)

The Attempt at a Solution

ln(y)= p(ln(p^x))x^p

1/y y'=p'(ln(p^x))x^p + p(logp)x^p + p(ln(p^x))px^(p-1) --use "relevant" equation

1/y y'= 0 + plogp(x^p) + p(ln(p^x))px^(p-1) -- simplify derivatives

y'= y(plogp(x^p) + p(ln(p^x))px^(p-1)) -- multiply both sides by y

y'= (p^(x^p))(x^p)(plogp(x^p) + p(ln(p^x))px^(p-1)) -- sub in y from beginning

Can someone point out where my error is, and how to fix it?

Thanks

Hello, ab94. Welcome to PF !

Is this \displaystyle y=(p^{x^p})(x^p)\ ?

Then you have ln(y) wrong.

 

[ab94]

Hello, ab94. Welcome to PF !

Is this \displaystyle y=(p^{x^p)}(x^p)\ ?

Then you have ln(y) wrong.

Yes, that's what it is. So I can't take ln of both sides and say

ln(y)=ln(p^x^p)
ln(y)= pln(p^x) ?

I thought We could bring down exponents when using that ln property.

 

[SammyS]

Yes, that's what it is. So I can't take ln of both sides and say

ln(y)=ln(p^x^p)
ln(y)= pln(p^x) ?

I thought We could bring down exponents when using that ln property.

No.

\displaystyle \ln(y)=\ln\left((p^{x^p})(x^p)\right)

Now use some properties of logs & differentiate.

 

[ab94]

No.

\displaystyle \ln(y)=\ln\left((p^{x^p})(x^p)\right)

Now use some properties of logs & differentiate.

so then would I split up the ln's into

ln(y)=ln(p^x^p) + ln(x^p) ?

If so I ended with y' = p^x^p(x^p) (plogp + p/x) ...don't think that's right

 

[SammyS]

so then would I split up the ln's into

ln(y)=ln(p^x^p) + ln(x^p) ?

If so I ended with y' = p^x^p(x^p) (plogp + p/x) ...don't think that's right

Show your steps, please.

 

[BloodyFrozen]

You almost got it.

How did you expand $ln(p^{x^{p}})$

After this, take the derivative.

A little late edit.

 

[ab94]

You almost got it.

$$ ln(y)~=~x^{p}\cdot ln(p)+p\cdot ln(x)$$

You see what I did there? It was the first term on the right side that you made an error.

After this, take the derivative.

Ohh thanks, I was taking the highest exponent (p) instead of the entire exponent (xp). I'll post my solution after I work it out

 

[BloodyFrozen]

Ohh thanks, I was taking the highest exponent (p) instead of the entire exponent (xp). I'll post my solution after I work it out

Good.

 

[ab94]

Ok so:

ln(y)= ln(p^x^p) + ln(x^p)
ln(y)=(x^p)ln(p) + (p)ln(x)
1/y y'= (px^(p-1))ln(p) + (x^p)/p + 0 + p/x

therfore y'= p^x^p(x^p)(px^(p-1))ln(p) + (x^p)/p + + p/x

Is this correct? or is there an error

 

[BloodyFrozen]

Ok so:

ln(y)= ln(p^x^p) + ln(x^p)
ln(y)=(x^p)ln(p) + (p)ln(x)
1/y y'= (px^(p-1))ln(p) + (x^p)/p + 0 + p/x

therfore y'= p^x^p(x^p)(px^(p-1))ln(p) + (x^p)/p + + p/x

Is this correct? or is there an error

You have the general idea, but your grouping is wrong.

$$\frac{dy}{dx} ~=~ y\cdot [px^{p-1}ln(x)+\frac{x^{p}}{p}+\frac{p}{x}]$$

 

[ab94]

You have the general idea, but your grouping is wrong.

$$\frac{dy}{dx} ~=~ y\cdot [px^{p-1}ln(x)+\frac{x^{p}}{p}+\frac{p}{x}]$$

Nvm I got the right answer, I used wolframalpha and I subtracted my answer from theirs, and got 0. (i fixed my own error up there ^)

Thanks for the help

 

[SammyS]

By the way: \displaystyle p=e^{\ln(p)}

Therefore,

\displaystyle y=(p^{x^p})(x^p)

\displaystyle <br /> =e^{\ln(p)x^p}(x^p)
​

Taking the derivative:

\displaystyle y&#039;=\ln(p)\,p\,x^{p-1}e^{\ln(p)x^p}(x^p)+e^{\ln(p)x^p}px^{p-1}

For this problem it looks like a better way than doing the logarithmic derivative.