# Homework Help: Determine dy/dx if y=(x^5)^lnx

1. Aug 11, 2015

### DevonZA

My attempt:

lny=ln(x^5)^lnx
lny=lnxlnx^5
1/y.dy/dx=(lnx)d/dx(lnx^5)+d/dx(lnx)(lnx^5)
= lnx(1/x^5)+1/x(lnx^5)
= (lnx/x^5)+(lnx^5/x) <---- not sure if this simplifies to 1/x^4+lnx^4?
dy/dx= y(lnx/x^5 + lnx^5/x)
= (x^5)^lnx(lnx/x^5 + lnx^5/x)

Any help would be appreciated.
D.

2. Aug 11, 2015

### RUber

So far so good
You forgot the chain rule. It should be:
$\ln x \frac{5x^4}{x^5} + \frac{\ln x^5 }{x}$
This will simplify nicely.

3. Aug 11, 2015

### SteamKing

Staff Emeritus
Strictly speaking, this is not a thread suitable for the Differential Equations forum.

4. Aug 12, 2015

### DevonZA

Thanks for your response RUber.
$\frac{d}{dx} lnx^5 = 5lnx^4$? Does it not? I am unsure how you got $\ln x \frac{5x^4}{x^5}$

5. Aug 12, 2015

### DevonZA

It has been moved, my apologies.

6. Aug 12, 2015

### pwsnafu

No. $\ln x^5 = 5 \ln x$.

7. Aug 12, 2015

### RUber

Let g(x) = x^5, then you have $\frac{d}{dx} \ln g(x) = \frac{1}{g(x)} g'(x)$ by the chain rule.

edit: Or you can take the simplification that pwsnafu gave first, and not have to worry about the chain rule.

8. Aug 12, 2015

### SteamKing

Staff Emeritus
You can't ignore those two letters "ln" in front of x5. They mean something special.

9. Aug 12, 2015

### Staff: Mentor

Not (as already mentioned).
$\frac{d}{dx} lnx^5 = \frac{d}{dx} 5 lnx = 5 \frac{d}{dx} lnx$ = ??
I am assuming that lnx^5 means ln(x5) and not (ln(x))5.
BTW, your original thread title was "Determine dy/dx of y=(x^5)^lnx". I changed "of" to "if" because you don't take "dy/dx" of something. This symbol already represents the derivative of y with respect to x. It is not a synonym of "take the derivative of".

10. Aug 12, 2015

### DevonZA

Okay let's see if I understand:

Determine $\frac{dy}{dx}$ if y=$(x^5)^{lnx}$

lny=ln$(x^5)^{lnx}$
lny=$lnxlnx^5$
$\frac{1}{y}$ $\frac{dy}{dx}$ = (lnx)$\frac{d}{dx}(lnx^5) +\frac{d}{dx}(lnx)(lnx^5)$
= $\ln x \frac{5x^4}{x^5} + \frac{\ln x^5 }{x}$
= ln5 + $lnx^4$
= $ln5x^4$
$\frac{dy}{dx}$ = $(x^5)^{lnx}$ $5lnx^4$
Mod note: edited the line above to fix the exponent.
I'm not confident in this answer but I've been trying all day and my brain hurts

Last edited by a moderator: Aug 12, 2015
11. Aug 12, 2015

### RUber

You are good to here...but you should not include the 5x^4/x^5 in the logarithm.
$\ln x \frac{5x^4}{x^5}$ means $\left( \frac{5x^4}{x^5} \right) \ln (x) = \frac{5\ln x}{x}$
$\frac{\ln x^5 }{x} =\frac{ 5\ln x}{x}$
$\ln (5x^4) \neq 5 \ln (x^4)$

12. Aug 12, 2015

### RUber

An alternate method would be to get the exponents out of the way early:
lny=ln$(x^5)^{lnx}$
lny=$lnxlnx^5 = 5 \ln x \ln x = 5 (\ln x) ^2$
$\frac{1}{y} \frac{dy}{dx}$ = $5 \frac{d}{dx}(\ln x)^2$
Which may be a more straightforward application with the chain rule.

13. Aug 12, 2015

### DevonZA

I am more confused now? I did what you said earlier:

14. Aug 12, 2015

### DevonZA

lny=ln$(x^5)^{lnx}$
lny=$lnxlnx^5 = 5 \ln x \ln x = 5 (\ln x) ^2$
$\frac{1}{y} \frac{dy}{dx}$ = $5 \frac{d}{dx}(\ln x)^2$
$\frac{dy}{dx}$ = $10(lnx).\frac{1}{x}$
x=1

15. Aug 12, 2015

### DevonZA

Should it be $\ln x . \frac{5x^4}{x^5}$
I am so confused

16. Aug 12, 2015

### RUber

This is right.
You forgot your 1/y on the left side.
I have no clue why you say x = 1.
$\frac1y \frac{dy}{dx}$ = $10(lnx).\frac{1}{x} =\frac{10\ln x }{x}$
$\frac{dy}{dx} = y \frac{10\ln x }{x}$
Then you are done.

17. Aug 12, 2015

### RUber

Yes. The chain rule says $\frac{d}{dx} \ln x^5 = \frac{1}{x^5} \times 5x^4$ Which is equal to $\frac5x$
This makes sense, since $\frac{d}{dx} \ln x^5 = \frac{d}{dx} (5\ln x) = 5 \frac{d}{dx} \ln x = \frac5x$

You had $\ln x \times \frac{d}{dx} \ln x^5$ which should be equal to $\ln x \times \frac{1}{x^5} \times 5x^4$

18. Aug 12, 2015

### DevonZA

Thank you RUber. Sorry I am slow I need more practice. Really appreciate your help.

19. Aug 12, 2015

### RUber

Not a problem. Take your time and look for chances to simplify before jumping into the derivative. Practice makes us all better.

20. Aug 16, 2015

### DevonZA

Final answer attached. Thanks to all who helped.

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