Question

Use logarithmic differentiation to find the derivative of the function. y = x^2x

Answer 1

Hi :)
When you get both bases and exponents as functions of x, usually, logarithmic differentiation is your only feasible choice:
\[\Large y = x^{2x}\]
Start by taking the ln of both sides...
\[\Large \ln(y) = \ln\left(x^{2x}\right)\]
can you take it from here?

Answer 2

Hi,
I am confused about how logarithms work. i understand how to normally find derivative, but not using ln functions

Answer 3

Well, there are two reasons to use the logarithmic functions:
One, it turns products into sums, which are much easier to differentiate. (generally)
and two, the log (or ln) itself has a rather simple derivative:
\[\Large \frac{d}{dt}\ln(t) = \frac1t\]

Answer 4

Have you ever heard of implicit differentiation?

Answer 5

Briefly...

Answer 6

Well, thankfully, this particular case is not a very complicated case of implicit differentiation.
In particular, with THIS in mind:
\[\Large \frac{d}{dt}\ln(t) = \frac1t\]
what is the derivative of the left side, with respect to x?
\[\Large \color{blue}{\ln(y) }= \ln\left(x^{2x}\right)\]

Answer 7

would it be 1/y

Answer 8

1/y, but since we're differentiating with respect to x (with implicit differentiation in mind) it better be
\[\Large \frac1y \cdot \frac{dy}{dx}\]or\[\Large \frac1y\cdot y'\]depending on how you were taught to differentiate implicitly....
So, catch me so far? ^_^

Answer 9

yes making sense

Answer 10

Okay, so much for the left-hand side.
Let's do the right, and differentiate it.
\[\Large \ln\left(x^{2x}\right)\]
Normally, when both the base and exponent are functions of x, it's differentiating is not straightforward, if not downright impossible...

However, since we have a log (ln), we can use one of its more beloved properties, THIS one...
\[\Large \ln b^p = p\ln b\]

Answer 11

IE, the exponent may 'go down' and simply be multiplied to the log... with that in mind, how would you 'deal' with \[\Large \ln\left(x^{2x}\right)\]

Answer 12

2xlnx?

Answer 13

You still here? Sorry, OS gave up on me T.T

You're right with 2x ln (x) by the way...
\[\Large 2x \ln(x)\]
Now, you can use the product rule to differentiate this, and also, once again recall that
\[\Large \frac{d}{dt}\ln(t) = \frac1t\]