check derivative

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September 24th 2006, 01:24 PM
Mathman24

check derivative

Hi could somebody please look through my answer to his homework assigment .

Given $z = sin(x + sin(t))$

show that $\frac{\partial z}{\partial x} \cdot \frac{\partial ^2 x}{\partial x \partial z} = \frac{\partial z}{\partial t} \cdot \frac{\partial ^2 z} {\partial x^2}$

By using the chain-rule I get:

$f_x(x,t) = cos(x + sin(1))$

$f_{xx}(x,t) = -sin(x + sin(1))$

$f_t(x,t) = cos(1) \cdot cos(x + sin(1))$

$f_{tt}(x,t) = 0$

Therefore

$\frac{\partial z}{\partial x} \cdot \frac{\partial ^2 x}{\partial x \partial z} = cos(x + sin(1)) \cdot 0 = cos(1) \cdot cos(x + sin(1)) \cdot 0 = \frac{\partial z}{\partial t} \cdot \frac{\partial ^2 z} {\partial x^2}$

Does this look okay?

Sincerely and God bless,

Fred
September 24th 2006, 01:37 PM
topsquark

Quote:

Originally Posted by Mathman24

Hi could somebody please look through my answer to his homework assigment .

Given $z = sin(x + sin(t))$

show that $\frac{\partial z}{\partial x} \cdot \frac{\partial ^2 x}{\partial x \partial z} = \frac{\partial z}{\partial t} \cdot \frac{\partial ^2 z} {\partial x^2}$

By using the chain-rule I get:

$f_x(x,t) = cos(x + sin(1))$

$f_{xx}(x,t) = -sin(x + sin(1))$

$f_t(x,t) = cos(1) \cdot cos(x + sin(1))$

$f_{tt}(x,t) = 0$

Therefore

$\frac{\partial z}{\partial x} \cdot \frac{\partial ^2 x}{\partial x \partial z} = cos(x + sin(1)) \cdot 0 = cos(1) \cdot cos(x + sin(1)) \cdot 0 = \frac{\partial z}{\partial t} \cdot \frac{\partial ^2 z} {\partial x^2}$

Does this look okay?

Sincerely and God bless,

Fred

Three things:
1. Latex is currently down.
2. Use instead of
3. Your derivatives are a bit off.

Given z = sin(x + sin(t))
(del z)/(del x) = cos(x + sin(t)) *1

not cos(x + sin(1)).

This is the only error I saw, though you kept doing it consistently. So your other partials are wrong as well.

-Dan
September 24th 2006, 01:52 PM
Mathman24

Hi

There then the partials must be

$f_{xx}(x,t) =-sin(x + sin(t))$

$f_t(x,t) = cos(t) \cdot cos(x + sin(t))$

$f_tt(x,t) = -sin(t) * cos(x+sin(t)) - (cos(t))^2 * sin(x+sint)).$

I have checked them myself, but do they look right to you now?

Sincerley Fred

p.s. my original reasoning regarding the chain rule, was that correct. I hope.?
September 24th 2006, 01:56 PM
topsquark

Quote:

Originally Posted by Mathman24

Hi

There then the partials must be

$f_{xx}(x,t) =-sin(x + sin(t))$

$f_t(x,t) = cos(t) \cdot cos(x + sin(t))$

$f_tt(x,t) = -sin(t) * cos(x+sin(t)) - (cos(t))^2 * sin(x+sint)).$

I have checked them myself, but do they look right to you now?

Sincerley Fred

p.s. my original reasoning regarding the chain rule, was that correct. I hope.?

It looks good now. :) In regard to your "PS" yes.

-Dan