Using Logarithmic differentiation, what would be the derivative of y = x^n

Question

zepdrix · Answer

Wait wait, why would we use logarithmic differentiation on this?
We can use the power rule.

Did you mean to post y=n^x?

zepdrix · Answer

Or it's just to get practice with Logarithmic Differentiation I guess? :)

anonymous · Answer

@zepdrix the professor said it could be solved that way..
and everytime I try solve, I get it wrong lol

zepdrix · Answer

Ah :)

anonymous · Answer

the answer is $$y'= y (\ln x +1)$$

anonymous · Answer

but I get$$y′=(n/x)y$$

zepdrix · Answer

Taking the natural log of each side gives us,
$$\large \ln y=\ln (x^n)$$

Applying a rule of logs allows us to write it like this,$$\large \ln y=n \ln x$$Taking the derivative gives us,
$$\large \frac{1}{y}y'=n\frac{1}{x}$$

Mmm yah looks like you're doing it correctly.
Hmm..

zepdrix · Answer

Oh.. I see what's going on.

zepdrix · Answer

The answer you have listed is the derivative of y=x^x.

zepdrix · Answer

$$y=x^x$$$$y'=y(\ln x+1)$$

zepdrix · Answer

So maybe teacher didn't match up the problems correctly or something :O

anonymous · Answer

ahhh, I guess so :l 
I think i solved it like 5 times, even with calculators, and I was a bit confused lol

thanks!

zepdrix · Answer

With the problem you were working on though, we can `confirm` that you're doing it correctly,$$\large y'=y\frac{n}{x} \qquad = \qquad (x^n)\frac{n}{x}\qquad=\qquad nx^{x-1}$$
Just by checking our work with the power rule :)

anonymous · Answer

thanks :)

mathstudent55 · Answer

$y = x^n$

$\ln y = \ln x^n$

$\ln y = n \ln x$

$\dfrac{y'}{y} = n\dfrac{1}{x} $

$y' = \dfrac{n}{x} y$