# Find the derivative of each implicit function, where dY/dX = -F_X/F_Y, provided that F_Y does not...

## Question:

Find the derivative of each implicit function, where {eq}\frac{dY}{dX} = -\frac{F_{X}}{F_{Y}} {/eq}, provided that {eq}F_{Y} \neq 0 {/eq}.

a) {eq}F(x,y) = x^{2} + y^{2} + (xy)^{\frac{1}{3}} = 0 {/eq}

b) {eq}F(x,y) = x^{2} y + y^{2}x + xy = 0 {/eq}

## Implicit Differentiation:

Implicit differentiation uses the chain rule to differentiate a function that is implicitly defined. Most of the problems are written as explicit functions of *x*. For example, {eq}y = 2x^2 + 4
{/eq}. For explicitly written functions, it is easy to find the derivatives of the function *y* with respect to the variable *x*. However, some functions in economics are written implicitly as functions of *x* and *y*. For example, {eq}U(x,y) = x^{1/3}y^{2/3}
{/eq}. For this function, the derivative of the function *y* with respect can only be differentiated using the chain rule.

## Answer and Explanation: 1

**a)** {eq}F(x,y) = x^{2} + y^{2} + (xy)^{\frac{1}{3}} = 0
{/eq}

The partial derivative of *F(x,y)* with respect to *x* is:

{eq}F_x = 2x + \frac{1}{3}(y)^{\frac{-2}{3}} {/eq}

The partial derivative of *F(x,y)* with respect to *y* is:

{eq}F_y = 2y + \frac{1}{3}(x)^{\frac{-2}{3}} {/eq}

Therefore:

{eq}\frac{dY}{dX} = -\frac{F_{x}}{F_{y}} = \displaystyle -\frac{2x + \frac{1}{3}(y)^{\frac{-2}{3}}}{2y + \frac{1}{3}(x)^{\frac{-2}{3}} } {/eq}

**b)** {eq}F(x,y) = x^{2} y + y^{2}x + xy = 0
{/eq}

The partial derivative of *F(x,y)* with respect to *x* is:

{eq}F_x = 2xy + y^2 + y {/eq}

The partial derivative of *F(x,y)* with respect to *y* is:

{eq}F_y = x^2 + 2xy + x {/eq}

Therefore:

{eq}\frac{dY}{dX} = -\frac{F_{x}}{F_{y}} = \displaystyle -\frac{2xy + y^2 + y}{x^2 + 2xy + x} {/eq}

#### Learn more about this topic:

from

Chapter 14 / Lesson 4This lesson defines the chain rule. It goes on to explore the chain rule with partial derivatives and integrals of partial derivatives.