Partial derivative with respect to x^2

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Yadavindu 2022-11-18

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评论： VBBV 2023-1-20

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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.

回答（3 个）

David Goodmanson 2022-11-18

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编辑：David Goodmanson 2022-11-18

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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.

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Yadavindu 2022-11-18

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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 2022-11-18

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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)

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KSSV 2022-11-18

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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)).

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