# Solve: {eq}f(x,y)=x^2-y^2,x^2+y^2=1 {/eq}

## Question:

Solve:

{eq}f(x,y)=x^2-y^2,x^2+y^2=1 {/eq}

## Partial Derivatives:

Let {eq}p(x, y) {/eq} be a function of two independent variables, then the partial derivative of a function of two variables can be found by applying the usual rules of differentiation with one exception. Since we have two variables, we must treat one as a constant when we take the derivative of the function with respect to the other variable.

That means that if we take the derivative with respect to {eq}y {/eq}, then the second variable {eq}x {/eq} is treated as a constant in every respect.

Similarly, when taking the derivative with respect to {eq}x {/eq}, the variable {eq}y {/eq} will be treated as a constant whenever it appears.

The following rules are relevant to this problem:

1.{eq}{\frac{d}{{dx}}\left( {f\left( x \right) + r\left( x \right)} \right) = \frac{d}{{dx}}f(x) + \frac{d}{{dx}}r(x)} {/eq}

2.{eq}{\frac{d}{{dx}}c = 0,\frac{d}{{dx}}{x^n} = n{x^{n - 1}}} {/eq}

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Given that: {eq}\displaystyle {x^2} + {y^2} = 1 {/eq}

{eq}\displaystyle \eqalign{ & {x^2} + {y^2} = 1 \cr & f\left( {x,y} \right) = {x^2} +...