Derivative of x^(lnx) & ln(cos-¹x)

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Homework Statement


y=x^(lnx) i know the derivative of lnx is 1/x but what about x^(lnx)?

ln(cos-¹x)


Homework Equations


N/A


The Attempt at a Solution



ln(cos-¹x) (1/(cos-¹x))(-x/√(1-x²)) -x/(cos-¹x√(1-x²))
The back in the book has the same answer except the -x is a -1 instead.
 
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Where did the cos-1x come from? :confused:

Usually, when you want to differentiate a function of the form f(x)^g(x), it helps to take the logarithm of the function and then find the derivative.

y = x^ln(x)

ln(y) = ln(x^ln(x))...
Can you do it from here?
 
on the ln(cos^-1x) is a different question.
 
Fusilli_Jerry89 said:
on the ln(cos^-1x) is a different question.
Oh, okay. There is a -1 because the derivative of arccos(x) is -1/sqrt(1-x^2).
 
k I got the first one but i don't get the second one.(ln(cos-¹x))
 
k nm I got it thx
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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