How to find the derivative of this?

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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
 

Answers and Replies

  • #2
SammyS
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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 [itex]\displaystyle y=(p^{x^p})(x^p)\ ?[/itex]

Then you have ln(y) wrong.
 
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  • #3
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Hello, ab94. Welcome to PF !

Is this [itex]\displaystyle y=(p^{x^p)}(x^p)\ ?[/itex]

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.
 
  • #4
SammyS
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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.

[itex]\displaystyle \ln(y)=\ln\left((p^{x^p})(x^p)\right)[/itex]

Now use some properties of logs & differentiate.
 
  • #5
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No.

[itex]\displaystyle \ln(y)=\ln\left((p^{x^p})(x^p)\right)[/itex]

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
 
  • #6
SammyS
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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.
 
  • #7
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You almost got it.

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

After this, take the derivative.

A little late edit.
 
  • #8
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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
 
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  • #9
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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.
 
  • #10
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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
 
  • #11
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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.

$$\frac{dy}{dx} ~=~ y\cdot [px^{p-1}ln(x)+\frac{x^{p}}{p}+\frac{p}{x}]$$
 
  • #12
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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
 
  • #13
SammyS
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By the way: [itex]\displaystyle p=e^{\ln(p)}[/itex]

Therefore,

[itex]\displaystyle y=(p^{x^p})(x^p)[/itex]
[itex]\displaystyle
=e^{\ln(p)x^p}(x^p)[/itex]

Taking the derivative:

[itex]\displaystyle y'=\ln(p)\,p\,x^{p-1}e^{\ln(p)x^p}(x^p)+e^{\ln(p)x^p}px^{p-1}[/itex]

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

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