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

$\frac{d}{dx} lnx^5 = 5lnx^4$? 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

$\frac{d}{dx} lnx^5 = \frac{d}{dx} 5 lnx = 5 \frac{d}{dx} lnx$ = ??
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.