# How to find partial derivatives in very abstract case when F is just F(x1,x2) and how to express it correctly?

I’ve just started learning differential calculus and there’s one task that I don’t completely understand. It sounds like:

“Given Y=F(x1,x2)+f(x1)+g(x2)Y=F(x_1,x_2)+f(x_1)+g(x_2), find ∂Y∂X1\frac {\partial Y}{\partial X_1}, ∂2Y∂X21\frac {\partial ^2 Y}{\partial X_1^2} and ∂2Y∂X1∂X2\frac {\partial ^2 Y}{\partial X_1 \partial X_2}”.
They don’t describe exactly what do the F, f and g functions look like.
Still, it’s quite easy to find the partial derivative for the f(x1)f(x_1) and g(x2)g(x_2) (∂Y∂X1f(x1)=f′(x1)\frac {\partial Y}{\partial X_1} f(x_1) = f'(x_1) and ∂Y∂X1g(x2)=0\frac {\partial Y}{\partial X_1} g(x_2) = 0 as x2=constx_2 = const.

But what to do with the “big” two-arguments function? What notation to use? Will
∂Y∂X1F(x,y)=∂Y∂X1F(x,y)x2\frac {\partial Y}{\partial X_1} F(x,y) = \frac {\partial Y}{\partial X_1} F(x,y)_{x_2} be just the required answer? (x2x_2 subscript means that we consider it a constant in this case)

But how to write the answers for ∂2Y∂X21\frac {\partial ^2 Y}{\partial X_1^2} and ∂2Y∂X1∂X2\frac {\partial ^2 Y}{\partial X_1 \partial X_2}? Will the result for ∂2Y∂X21\frac {\partial ^2 Y}{\partial X_1^2} look just like the one above? And what for the ∂2Y∂X1∂X2\frac {\partial ^2 Y}{\partial X_1 \partial X_2}?

I’ve tried WolframAlpha just to give me some clues, but it uses some notation I don’t understand:

∂Y∂X1F(x,y)=F(1,0)(x,y)\frac {\partial Y}{\partial X_1} F(x,y) = F^{(1,0)}(x,y)

I’m getting quite confident when it comes to differentiating some “concrete” functions but I somehow get stuck when it comes to such abstract situations.

Will appreciate any help,

Thanks,

Paul

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Y=F(x1,x2)+f(x1)+g(x2)⟹∂∂x1Y=∂∂x1[F(x1,x2)+f(x1)+g(x2)]⟹∂Y∂x1=∂F(x1,x2)∂x1+∂f(x1)∂x1+∂g(x2)∂x1=∂F(x1,x2)∂x1+df(x1)dx1+0=∂F(x1,x2)∂x1+df(x1)dx1Y=F(x_1,x_2)+f(x_1)+g(x_2) \\ \implies \frac{\partial}{\partial x_1} Y = \frac{\partial}{\partial x_1} \left[F(x_1,x_2)+f(x_1)+g(x_2)\right] \\ \begin{align}\implies \require{enclose}\enclose{box}{\frac{\partial Y}{\partial x_1}} &= \frac{\partial F(x_1,x_2)}{\partial x_1}+ \frac{\partial f(x_1)}{\partial x_1}+\frac{\partial g(x_2)}{\partial x_1} \\ &= \frac{\partial F(x_1,x_2)}{\partial x_1}+ \frac{df(x_1)}{dx_1}+0 \\ &\enclose{box}{= \frac{\partial F(x_1,x_2)}{\partial x_1}+ \frac{df(x_1)}{dx_1}}\end{align}

And so for the second derivative, say, with respect to x2, the F(x1,x2) part will be d^2F(x1,x2) in the numerator and dx1dx2 in the denominator (d will be the partial derivative signs), while the second part (for f(x1)) will be 0, right?
– sempol
2 days ago

(it seems I can use MathJax here too, so I’ll rewrite it for clarity) the F(x_1,x_2)F(x_1,x_2) part will be \frac {\partial^{2}F(x_1,x_2)}{\partial x_1 \partial x_2}\frac {\partial^{2}F(x_1,x_2)}{\partial x_1 \partial x_2}, while the second part will be 0 (so =frac {\partial^{2}F(x_1,x_2)}{\partial x_1 \partial x_2}=frac {\partial^{2}F(x_1,x_2)}{\partial x_1 \partial x_2}
– sempol
2 days ago

That is correct.
– Bye_World
2 days ago