Using Chain Rule to Find Partial Derivatives of a Multivariable Function

In summary, the chain rule was used to find the partial derivatives of z with respect to s and t. The correct equations for the partial derivatives were substituted into the formula, but there was an error in the last term where the squared term was missing. After correcting this, the final answer was obtained.
  • #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.
 
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  • #2
Turbodog66 said:

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)##.
 
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  • #3
Ray Vickson said:
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?
 

FAQ: Using Chain Rule to Find Partial Derivatives of a Multivariable Function

What is the Chain Rule?

The Chain Rule is a mathematical rule that allows you to find the derivative of a composite function. It is used to calculate the rate of change of a quantity that depends on multiple variables.

How do you apply the Chain Rule?

To apply the Chain Rule, you need to first identify the composite function and then take the derivative of the outer function multiplied by the derivative of the inner function.

Why is the Chain Rule important?

The Chain Rule is important because it allows us to find the derivative of complex functions by breaking them down into simpler functions. This is crucial in many areas of science, such as physics and engineering, where functions can be very complex and difficult to differentiate.

Can the Chain Rule be used for higher order derivatives?

Yes, the Chain Rule can be applied to higher order derivatives. It follows the same principle of taking the derivative of the outer function multiplied by the derivative of the inner function.

What are some common applications of the Chain Rule?

The Chain Rule has many applications in science and engineering, such as in physics for finding the acceleration of an object, in economics for calculating marginal costs, and in biology for understanding growth rates of populations.

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