1. Sep 24, 2006

mathboy20

Hi

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 that look right ?

Sincerely
MM20

Last edited: Sep 24, 2006