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**Mathman24** Hi could somebody please look through my answer to his homework assigment .

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

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

By using the chain-rule I get:

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

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

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

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

Therefore

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

Does this look okay?

Sincerely and God bless,

Fred