Chain Rule

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  • #1
Turbodog66
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Homework Statement


Suppose $$z=x^2 sin(y), x=5t^2-5s^2, y=4st$$
Use the chain rule to find $$\frac{\partial z}{\partial s} \text{ and } \frac{\partial z}{\partial t}$$

Homework Equations


$$\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial s}$$

The Attempt at a Solution


$$\frac{\partial z}{\partial x} = 2x\sin(y)$$
$$\frac{\partial x}{\partial s} = -10s$$
$$\frac{\partial z}{\partial y} = x^2\cos(y)$$
$$\frac{\partial y}{\partial s} = 4t$$
With those I then substitute the values into the equation, and I came up with $$\frac{\partial z}{\partial s}=2x\sin(y)(-10s)+x^2\cos(y)(4t)=-20(5t^2-5s^2)(s)\sin(4st)+4(5t^2-5s^2)(t)\cos(4st)$$
Where am I going wrong? Also, I am still learning Latex, so I apologize for the crude display.
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


Suppose $$z=x^2 sin(y), x=5t^2-5s^2, y=4st$$
Use the chain rule to find $$\frac{\partial z}{\partial s} \text{ and } \frac{\partial z}{\partial t}$$

Homework Equations


$$\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial s}$$

The Attempt at a Solution


$$\frac{\partial z}{\partial x} = 2x\sin(y)$$
$$\frac{\partial x}{\partial s} = -10s$$
$$\frac{\partial z}{\partial y} = x^2\cos(y)$$
$$\frac{\partial y}{\partial s} = 4t$$
With those I then substitute the values into the equation, and I came up with $$\frac{\partial z}{\partial s}=2x\sin(y)(-10s)+x^2\cos(y)(4t)=-20(5t^2-5s^2)(s)\sin(4st)+4(5t^2-5s^2)(t)\cos(4st)$$
Where am I going wrong? Also, I am still learning Latex, so I apologize for the crude display.

In the last term you should have ##(5t^2-5s^2)^2##, not ##(5t^2-5s^2)##.
 
  • #3
Turbodog66
13
0
In the last term you should have ##(5t^2-5s^2)^2##, not ##(5t^2-5s^2)##.
Thanks! Never fails I overlook something simple like that. Other than that, does it appear that I am going about it correctly?
 

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