# Thread: Indice differentiation

1. ## Indice differentiation

Given
$f(x,y)=x^{xy}$

How do we evaluate:
${{\partial f} \over {\partial x}}$
and
${{\partial f} \over {\partial y}}$

I generated the answer with mathematica and the answers seem different from a textbook.
Thanks

2. ## Clarification?

chopet, what do you mean by the comma in the exponent?

--Kevin C.

3. no comma, sorry. Plain xy to be differentiated.

4. Originally Posted by chopet
Given
$f(x,y)=x^{xy}$

How do we evaluate:
${{\partial f} \over {\partial x}}$
and
${{\partial f} \over {\partial y}}$

I generated the answer with mathematica and the answers seem different from a textbook.
Thanks
$f(x,y) = x^{xy} = \left ( x^x \right ) ^y$

So
$\frac{\partial f}{\partial x} = y \left ( x^x \right )^{y - 1} \cdot (ln(x) + 1) \cdot x^x$
by the chain rule so
$\frac{\partial f}{\partial x} = y(ln(x) + 1) \left ( x^x \right )^y$

Since x is a constant with respect to the y derivative:
$\frac{\partial f}{\partial y} = ln \left ( x^x \right ) \cdot \left ( x^x \right ) ^y$

$\frac{\partial f}{\partial y} = x \cdot ln(x) \cdot \left ( x^x \right ) ^y$

-Dan