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Posted by **Carolina** on Tuesday, October 13, 2009 at 5:45pm.

y= x^(lnx)

thanks

- Calc--derivatives -
**MathMate**, Tuesday, October 13, 2009 at 6:01pmUse 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.

- Calc--derivatives -
**Carolina**, Tuesday, October 13, 2009 at 6:07pmummm.. i dont understand your language!

can you explain that in baby terms? thanks!

- Calc--derivatives -
**Reiny**, Tuesday, October 13, 2009 at 6:10pmhow 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.

- Calc--derivatives -
**Carolina**, Tuesday, October 13, 2009 at 6:11pmdid u get

e^(lnx) / x^2 ??

- Calc--derivatives -
**MathMate**, Tuesday, October 13, 2009 at 6:12pmI 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).

- Calc--derivatives -
**MathMate**, Tuesday, October 13, 2009 at 6:17pmFor 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.