Question

find the der of

y= x^(lnx)

thanks

Answer 1

Use the function of a function rule, namely
dy/dx = dy/dy . du/dx
substitute u=log(x)
and find the derivative of
d(x^u)/du then multiply by du/dx.

Answer 2

ummm.. i don't understand your language!

can you explain that in baby terms? thanks!

Answer 3

how about this
take ln of both sides

ln y = ln(x^lnx)
ln y = lnx(lnx) = (lnx)^2

now differentiate implicitly
(dy/dx)/y = 2lnx * 1/x
dy/dx = y(2lnx)/x
= (x^lnx) (2lnx)/x

somebody please check this, I have been making some silly mistakes today.

Answer 4

did u get

e^(lnx) / x^2 ??

Answer 5

I will try with an example.

Use the function of a function rule, namely
dy/dx = dy/dy . du/dx
substitute u=log(x)
and find the derivative of
d(x^u)/du then multiply by du/dx.

Suppose y=sin(x²)
We know how to differentiate sin(x), and how to do x², but we're not sure how to find the derivative of sin(x²).

So let u=x²
then
y=sin(u)
dy/du = cos(u)
We also know that
du/dx = d(x²)/dx = 2x

So, by the chain rule,
dy/dx = dy/du.du/dx = 2xcos(x²)

You can apply the same procedure to your problem using u=ln(x).

Answer 6

For Carolina,
There is no e^x in the answer.
The derivative of log(x) is 1/x.

Also,
Reiny's answer is right.
Mine comes up to
2*x^(log(x)-1)*log(x)
which is the same thing.