Partial derivative with respect to x^2

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.

Answer 1

[David Goodmanson]

Hi Yadavindu,

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.

[Yadavindu]

Thank you for your reply, so we don't have to approach the question directly rather we have approach in this way,

syms x y z
f= (x^5*y^3+ z^2*x^2+y*x) / (x^3*y+x*y)

f = 

f1=diff(f,x)

f1 = 

f2= f1/2*x

f2 = 

[David Goodmanson]

not quite, you meant to say

f2 = f1/(2*x)

[Yadavindu]

Yes

[Yadavindu]

so the code would be

syms x y z
f= (x^5*y^3+ z^2*x^2+y*x) / (x^3*y+x*y)

f = 

f1=diff(f,x)

f1 = 

f2= f1/(2*x)

f2 = 

[David Goodmanson]

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)

Answer 2

[KSSV]

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 = 

[Yadavindu]

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)).

Answer 3

[VBBV]

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]

you can assume the x^2 as t and differentiate with t , then back substitute with x^2

[Yadavindu]

Thank you for your reply

[VBBV]

If it solved your problem (which i hope it did) pls accept the answer.