Differentiating a Symbolic Function

If I have the following symbolic function:

syms f(x,y)
f(x,y) = 2*x^2 + y;

When I try to diff. this function w.r.t. (x) I get:

d = diff(f,x)
d(x, y) =
4*x

Which is a function in both x, and y NOT in (x) only.

How can I perform this and get the real exsisting function input only "i.e. (x) only here"?

Answer 1

[Star Strider]

That is the corrct result, since diff is taking the partial derivative with respect to ‘x’ only, and ‘y’ is considered a constant.

To get the derivatives of ‘f’ with respect to both variables, use the jacobian function:

jf = jacobian(f)

producing:

jf(x, y) =

[4*x, 1]

See Differentiation for more information.

Although not appropriate here, explore the functionalDerivative function.

[Ammar Taha]

First of all thanks for your time. Second, I know that this is a correct answer. What I meant is that for the derivatve function instead of having it as a function of (x), and (y) I need it to be a function of the variable I diff. w.r.t. only, so that when I want to evaluate that derivative function I have to plug in (x) only "rather (x) and 0 for (y)".

[Star Strider]

My pleasure!

Note that ‘d’ is a funciton of both ‘x’ and ‘y’, even though ‘y’ does not exist in it:

d(x, y) =
4*x

The result is that:

v = d(2,3)

produces:

v =
8

and no errors, while:

v = d(2)

produces:

Error using symfun/subsref (line 189)
Invalid argument at position 2. Symbolic function expected 2 input arguments but received 1.

However if you define it as:

d(x) = diff(f,x)

then:

v = d(2)

produces the expected result and no errors.

[Ammar Taha]

Ok, thats great. Thnaks a lot.

[Star Strider]

As always, my pleasure!

Answer 2

[Chendi Lin]

Hi Ammar,

Without explicitly defining the differentiation variable, "diff" uses the default variable, which is "x" in your case.

To get both derivatives, you can do

[diff(f(x,y),x) diff(f(x,y),y)]

And this should give you the correct result.

Please refer to this document: Differentiate symbolic expression or function - MATLAB diff (mathworks.com)

Thanks.

CD

[Ammar Taha]

Ok, thanks a lot.