Urgend Calculus Question: Please Look

[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

 

[mighty2000]

Hello MM20,

Yes, your approach and solution are correct. You have correctly used the chain rule to find the first and second derivatives of z with respect to x and t. Then, by plugging these values into the given equation, you have shown that the two sides are equal. This demonstrates that the equation is valid and holds true for the given function z. Well done! Keep up the good work and keep exploring mathematical concepts.