Partial derivative question (2 Viewers)

integral95 · Aug 6, 2014

Ok I maybe I'm derping and stuff but i can't seem to get this one

$$Suppose that function f is a differentiable function of a single variable and F(x,y) is defined by$ \ \ F(x,y) = \ f(x^2-y) \\ $a) Show that$ \frac{\partial F}{\partial x} + 2x\frac{\partial F}{\partial y} = 0 \\ \\ b) \ \ \ $Given that$ \ \ F(0,y) = sin y,\ \ $find a formula for $F(x,y)$

The answer is $F(x,y) = sin(y-x^2)$

braintic · Aug 6, 2014

What is the significance of small f ?

Where does it appear in the question?

And ... haven't you already given a formula for F(x,y) ?

Sy123 · Aug 6, 2014

integral95 said:
Ok I maybe I'm derping and stuff but i can't seem to get this one

$$Suppose that function f is a differentiable function of a single variable and F(x,y) is defined by$ \ \ F(x,y) = f(x^2-y) \\ $Given that$ \frac{\partial F}{\partial x} + 2x\frac{\partial F}{\partial y} = 0 \\ \\ and F(0,y) = sin y, $find a formula for $F(x,y)$

EDIT: ignore that br> stuff, i dunno why it's happen sorry i'm not that good with latex

The br stuff appeared since you finished the line with pressing enter

Is the above what you meant to ask?

What does little f mean?

integral95 · Aug 7, 2014

Sy123 said:
The br stuff appeared since you finished the line with pressing enter

Is the above what you meant to ask?

What does little f mean?

Post edited!..... that's just how the question is given.

integral95 · Aug 7, 2014

Sy123 · Aug 7, 2014

integral95 said:
Ok I maybe I'm derping and stuff but i can't seem to get this one

$$Suppose that function f is a differentiable function of a single variable and F(x,y) is defined by$ \ \ F(x,y) = \ f(x^2-y) \\ $a) Show that$ \frac{\partial F}{\partial x} + 2x\frac{\partial F}{\partial y} = 0 \\ \\ b) \ \ \ $Given that$ \ \ F(0,y) = sin y,\ \ $find a formula for $F(x,y)$

The answer is $F(x,y) = sin(y-x^2)$

$\\ $By partially differentiating, we are simply differentiating the function with respect to the variable treating all else as constant$ \\ \\ $So$ \ \frac{\partial F}{\partial x} = \frac{d}{dx} f(x^2-y) = 2x f'(x^2-y) \\ \\ \frac{\partial F}{\partial y} = -f'(x^2-y)$

$\\ $It is clear from these 2 equations, by multiplying the second by$ \ 2x \ $and adding to the first, the result is zero$$

$\\ $Part B:$ \\ \\ $Since$ \ F(0,y) = \sin y \\ \\ $Then since$ \ F(0,y) = f(0^2 - y) = f(-y) = \sin y \\ \\ \therefore \ f(x) = -\sin x$

$\\ $So,$ \ f(x^2-y) = \sin(y-x^2) = F(x,y) \ \square$

Partial derivative question (2 Viewers)

integral95

Well-Known Member

braintic

Well-Known Member

Sy123

This too shall pass

integral95

Well-Known Member

integral95

Well-Known Member

Sy123

This too shall pass

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