Posted by Carolina on Tuesday, October 13, 2009 at 5:45pm.
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.
ummm.. i dont understand your language!
can you explain that in baby terms? thanks!
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.
did u get
e^(lnx) / x^2 ??
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).
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.
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