# Partial derivative with respect to x^2

5 views (last 30 days)
Commented: VBBV on 20 Jan 2023 at 22:47
Suppose I have a function f
f = (x^5*y^3+ z^2*x^2+y*x) / (x^3*y+x*y)
how do I take derivative of this function with respect to x^2.
I have used diff(f, x^2) but it is returning an error.
syms x y z
f= (x^5*y^3+ z^2*x^2+y*x) / (x^3*y+x*y)
f = diff(f,x^2)
Error using sym/diff
Second argument must be a variable or a nonnegative integer specifying the number of differentiations.

David Goodmanson on 18 Nov 2022
Edited: David Goodmanson on 18 Nov 2022
df/d(x^2) = (df/dx) / (d(x^2)/dx) = (df/dx) / (2*x)
which you can code up without the issue you are seeing.
David Goodmanson on 18 Nov 2022
yes, although you could write it in one line and toss in a simplify:
syms x y z
f= (x^5*y^3+ z^2*x^2+y*x) / (x^3*y+x*y);
f1 = simplify(diff(f,x)/(2*x))
f1 = (2*x^5*y^3 + 4*x^3*y^3 - x^2*z^2 - 2*x*y + z^2)/(2*x*y*(x^2 + 1)^2)

KSSV on 18 Nov 2022
syms x y z
f= (x^5*y^3+ z^2*x^2+y*x) / (x^3*y+x*y)
f = dfdx = diff(f,x)
dfdx = dfdx2 = diff(dfdx,x)
dfdx2 = Thank you for your reply, however I think, if I take derivative of d(df\dx)\dx = (d^2(f) \ (dx)(dx)) will give double partial derivative rather my question was whether there exist any direct way to calculate (df / d(x^2)).

VBBV on 18 Nov 2022
Edited: VBBV on 18 Nov 2022
syms x y z t
f= (x^5*y^3+ z^2*x^2+y*x) / (x^3*y+x*y)
f = % t = x^2 % assume x = sqrt(t)
F = subs(f,x,sqrt(t))
F = y = diff(F,t)
y = Y = subs(y,t,x^2) % back substitute with x
Y = VBBV on 20 Jan 2023 at 22:47
If it solved your problem (which i hope it did) pls accept the answer.

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