How to take the Partial Derivatives of a Function that is Defined Implicitly?

In summary, to take the partial derivatives of a function that is defined implicitly, one must specify the function first. In the given example, the equation x^2 / 4 + y^2 + z^2 = 3 can be thought of as defining z implicitly in terms of x and y. The partial derivatives of z with respect to x and y can be found by differentiating the equation and solving for z_x and z_y. However, since x and y are independent variables, their derivatives with respect to each other are 0. Alternatively, the equation can also be thought of as defining y or x in terms of the other variables, resulting in different partial derivatives.
  • #1
jaguar7
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How does one take the partial derivatives of a function that is defined implicitly? For example, the function, x^2 / 4 + y^2 + z^2 = 3.
 
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  • #2
To take the partial derivatives of a function, you first have to specify the function!
[tex]\frac{x^2}{4}+ y^2+ z^2= 3[/tex]
is not a function, it is an equation. If you mean that equation defines z implicitely as a function of x and y, then, differentiating with respect to x, [itex]x/2+ 2z z_x= 0[/itex] so that [itex]z_x= -x/4z[/itex] and [itex]2y+ 2z z_y= 0[/itex] so that [itex]z_y= -y/z[/itex]. Notice that, in the first case, the derivative of y with respect to x is 0 and, in the second, the derivative of x with resepect to y is 0. That is because they are independent variables.

Of course, we could just as easily think of that equation as defining y in terms of the variables x and z and have [itex]x/2+ 2yy_x= 0[/itex] and [itex]2y y_z+ 2z= 0[/itex] or we could think of it as defining x in terms of y and z so that [itex](x/2)x_y+ 2y= 0[/itex] and [itex](x/2)x_z+ 2z= 0[/itex].
 

Related to How to take the Partial Derivatives of a Function that is Defined Implicitly?

1. What is an implicit function?

An implicit function is a function that is defined implicitly, meaning that it is not explicitly written in terms of its independent variables. Instead, the function is defined through an equation that relates the dependent and independent variables.

2. Why do we need to take the partial derivatives of an implicit function?

Partial derivatives of an implicit function are needed to find the rate of change of the function with respect to each of its independent variables. This is important in many fields of science, such as physics, economics, and engineering, where understanding the behavior of a system requires knowing how it changes in response to different variables.

3. What is the process for taking the partial derivatives of an implicit function?

The process for taking the partial derivatives of an implicit function involves treating the other independent variables as constants and using the chain rule to differentiate the function with respect to the variable of interest. This results in an equation for the partial derivative in terms of the other variables and the original equation defining the implicit function.

4. Can you provide an example of taking the partial derivatives of an implicit function?

Sure, let's consider the implicit function x^2 + y^2 = 25. To find the partial derivative with respect to x, we treat y as a constant and differentiate both sides of the equation with respect to x. This gives us the equation 2x + 2yy' = 0, where y' is the partial derivative of y with respect to x. Solving for y', we get y' = -x/y. Similarly, the partial derivative with respect to y can be found by treating x as a constant, giving us the equation 2y + 2xy' = 0 and solving for y', resulting in y' = -y/x.

5. What are some applications of implicit functions and their partial derivatives?

Implicit functions and their partial derivatives have many applications in various fields of science and engineering. For example, they are used in thermodynamics to determine the relationships between different thermodynamic variables, in economics to analyze consumer behavior, and in fluid mechanics to understand the behavior of fluids in complex systems. They are also crucial in optimization problems, where finding the maximum or minimum of a function requires taking partial derivatives and setting them equal to zero.

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