Extremely hard calculus question.

**integral** · June 23rd 2010, 11:18 AM

Extremely hard for me at least!
This problem is just something I came up with. for practicing properties, and how to use partial.

I just want to know if I did this right.. Sorry, but no one I know can do this stuff, not even the cal teacher at my higschool

.

consider: $t=x^{x^y}$
Find $\frac{\partial t_x}{\partial x}$ and
Then, say $\frac{\partial t_x}{\partial x}=z_x$

Find $\nabla z_x$

I found (hopefully correctly) that $z_x=x^{x^y}\left [ yx^{y-1} \right ] ln(x)+x^{y-1}$

Then
$\nabla z=\frac{\partial z_x}{\partial x}\mathbf{i}+\frac{\partial z_y}{\partial y}\mathbf{j}}$
$z=\left [x^{x^y} y\right]\left [x^{y-1} ln(x)\right ]+x^{y-1}$

$\frac{\partial z}{\partial x}=\frac{\partial}{\partial x}\left [x^{x^y} y\right]\left [x^{y-1} ln(x)\right ]+\left [x^{x^y} y\right]\frac{\partial}{\partial x}\left [x^{y-1} ln(x)\right ]+(y-1)x^{y-2}$

$\frac{\partial z}{\partial x}=\left [ {yz_x} \right ]\left [ x^{y-1}ln(x) \right ]+ \left [ \left [ (y-1)x^{y-2}ln(x) \right ]+x^{y-2} \right ]yx^{x^y}+(y-1)x^{y-2}$
$=\left [y \left [x^{x^y}\left [ yx^{y-1} \right ] ln(x)+x^{y-1} \right] \right ]\left [ x^{y-1}ln(x) \right ] + \left [ \left [ (y-1)x^{y-2}ln(x) \right ]+x^{y-2} \right ] yx^{x^y}+(y-1)x^{y-2}$

$\frac{\partial z}{\partial x}\mathbf{i}=\left [\left [\left [x^{x^y}\left [ yx^{y-1} \right ] ln(x^y)+yx^{y-1} \right] \right ]\left [ x^{y-1}ln(x) \right ] + \left [ \left [ (y-1)x^{y-2}ln(x) \right ]+x^{y-2} \right ] yx^{x^y}+(y-1)x^{y-2}\right ] \mathbf{i}$
Is that right?

I can't figure out the partial with respect to y though.. Can someone help?

**p0oint** · June 23rd 2010, 11:36 AM

$h=x^{x^y}$

$ln(h)=x^yln(x)$

$\frac{1}{h}h'=(x^yln(x))ln(x)$

$h'=x^{x^y}x^yln^2(x)$

$h'=x^{x^y+y}ln^2(x)$

Regards.

**Ackbeet** · June 23rd 2010, 11:45 AM

I think p0oint might be a little confused. t is a function of x and y. It is not true that y=x^(x^y).

I agree with integral's $z_{x}$ computation. As for $\frac{\partial x}{\partial x}$ , I get

$x^{-2 + x^y + y}\,\left( -1 + x^y + 2\,y + y\,\ln (x)\,\left( -1 + y + {x}^y\,\left( 2 + y\,\ln (x) \right) \right) \right)$ . You can check to see if it matches up with your answer.

Can you start on the y derivative?

**p0oint** · June 23rd 2010, 11:51 AM

Sorry, it was notation problem.

Now I changed it to h(x) which is the partial derivative of t with respect to y.

Regards.

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