# Partial derivative with respect to x^2

조회 수: 6(최근 30일)
댓글: VBBV 2023년 1월 20일 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.

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### 답변(3개)

David Goodmanson 2022년 11월 18일
편집: David Goodmanson 2022년 11월 18일
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.
##### 댓글 수: 5표시숨기기 이전 댓글 수: 4
David Goodmanson 2022년 11월 18일
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 2022년 11월 18일
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 = ##### 댓글 수: 1표시숨기기 없음
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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VBBV 2022년 11월 18일
편집: VBBV 2022년 11월 18일
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 = ##### 댓글 수: 3표시숨기기 이전 댓글 수: 2
VBBV 2023년 1월 20일 22:47
If it solved your problem (which i hope it did) pls accept the answer.

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