# Find the derivative with respect to x for f(x, y)= x^y.

## Question:

Find the derivative with respect to x for {eq}f(x, y)= x^y. {/eq}

## Partial Derivatives:

Let {eq}q(a, b) {/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.

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

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

The partial derivatives of {eq}\displaystyle q(a,b)\,\,{\text{at}}\,\,(x,y) {/eq} are defined as

{eq}{\left( {\frac{{\partial q}}{{\partial a}}} \right)_{\left( {x,y} \right)}} = \mathop {\lim }\limits_{k \to 0} \frac{{q(x + k,y) - h(x,y)}}{k} {/eq}

The following rules are relevant to this problem:

1.{eq}{\frac{d}{{dx}}{x^n} = n{x^{n - 1}}} {/eq}

2.{eq}{a + b = 0\,\, \Rightarrow a = - b} {/eq}

Given that: {eq}\displaystyle f(x,y) = {x^y} {/eq}

{eq}\displaystyle\ \eqalign{ & f(x,y) = {x^y} \cr & {\text{Differentiating }}f(x,y)\,\,{\text{w}}{\text{.r}}{\text{.t}}{\text{. '}}x{\text{';}} \cr & \frac{{\partial f}}{{\partial x}} = \frac{\partial }{{\partial x}}{x^y} \cr & {\text{From power formula;}} \cr & \,\,\,\,\,\,\,\, = y{x^{y - 1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\frac{d}{{dx}}{x^n} = n{x^{n - 1}}} \right) \cr & \cr & \frac{{\partial f}}{{\partial x}} = y{x^{y - 1}} \cr} {/eq}