Good afternoon all. If someone would kindly answer the following questions, it would be greatly appreciated. Suppose I have a function
f[x_,y_] := x2 + y2 + xy2 ...............................................................(1)
I can take the partial derivative with respect to x as follows:
D[f[x,y],x]........................................................................................(2)
and the derivative with respect to y as follows:
D[f[x,y],x]........................................................................................(3)
How can I request the partial derivative of x with respect to y ? This is the ratio of the result from (2) divided by the result from (3) with a minus sign. Is there a direct way to do it starting with the function itself?
BTW, the function is implicit (f(x,y) is zero. I'm not sure if that makes a difference.
Best regards,
Hazem
Good afternoon, Hazem. The partial derivative of xxx with respect to yyy, given that the function f(x,y)f(x,y)f(x,y) is implicitly defined by f(x,y)=0f(x,y) = 0f(x,y)=0, can indeed be computed directly. This type of derivative is known as the implicit derivative.
When f(x,y)=0f(x, y) = 0f(x,y)=0, the relationship between xxx and yyy is implicit, and to find ∂x∂y\frac{\partial x}{\partial y}∂y∂x, we use the concept of implicit differentiation.
Given the function: f(x,y)=x2+y2+xy2f(x, y) = x^2 + y^2 + xy^2f(x,y)=x2+y2+xy2
The implicit derivative ∂x∂y\frac{\partial x}{\partial y}∂y∂x can be found using the formula: ∂x∂y=−∂f∂y∂f∂x\frac{\partial x}{\partial y} = -\frac{\frac{\partial f}{\partial y}}{\frac{\partial f}{\partial x}}∂y∂x=−∂x∂f∂y∂f
Substituting the partial derivatives from above: ∂x∂y=−2y+2xy2x+y2\frac{\partial x}{\partial y} = -\frac{2y + 2xy}{2x + y^2}∂y∂x=−2x+y22y+2xy
Thus, the implicit derivative of xxx with respect to yyy for the function f(x,y)=x2+y2+xy2f(x, y) = x^2 + y^2 + xy^2f(x,y)=x2+y2+xy2 is: ∂x∂y=−2y+2xy2x+y2\frac{\partial x}{\partial y} = -\frac{2y + 2xy}{2x + y^2}∂y∂x=−2x+y22y+2xy
In summary, using implicit differentiation, we directly obtained the desired result: ∂x∂y=−∂f∂y∂f∂x\frac{\partial x}{\partial y} = -\frac{\frac{\partial f}{\partial y}}{\frac{\partial f}{\partial x}}∂y∂x=−∂x∂f∂y∂fThis method works starting from the function itself without needing to separately compute and divide the results of individual partial derivatives.
I don't know if my answer can be helpful for you, I asked your question to chat GPT and copy and paste its answer, let me know
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