MHB How Are Partial Derivatives Calculated for Multivariable Functions?

[kupid]

Its about functions with two or more variables ?

Partial Derivatives in Calculus

Let f(x,y) be a function with two variables.

If we keep y constant and differentiate f (assuming f is differentiable) with respect to the variable x, we obtain what is called the partial derivative of f with respect to x

Similarly

If we keep x constant and differentiate f (assuming f is differentiable) with respect to the variable y, we obtain what is called the partial derivative of f with respect to y

How do you keep this x and y constant , i don't understand .

 

[I like Serena]

Its about functions with two or more variables ?

How do you keep this x and y constant , i don't understand .

Hi kupid,

It means we treat them as constants, such as $3$.
Recall that the derivative of a constant is 0.
So if we consider x to be constant and differentiate it with respect to y, the result is 0.

To give a couple of examples:
$$\pd{}y x = 0 \\ \pd{}y (3x^2y) = 3x^2 \\ \pd{}y (xy^2) = 2xy$$

In reality x might actually be a function of y, but that's the difference between a total derivative and a partial derivative.
For a partial derivative we treat x just like any other constant, and its derivative is 0.

 

[kupid]

I like Serena ,

Thanks a lot :-)

From here is it possible to understand partial differential equations ?

 

[I like Serena]

I like Serena ,

Thanks a lot :-)

From here is it possible to understand partial differential equations ?

Well, that's a step up from just understanding what a partial derivative is.
But sure, why not.As an example, suppose we have the function $u(x,y)$.
And the partial differential equation
$$\pd{}x u(x,y)=0$$
Since we'd treat $y$ as a constant, the solution is some function of $y$.
That is, the solution is:
$$u(x,y) = f(y)$$
where $f$ is any arbitrary function of $y$.

We can see that if we take the partial derivative with respect to x, we will indeed get 0.

 

[kupid]

Thanks