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Found in: Page 658

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Equation 6 is a formula for the derivative $$\frac{{dy}}{{dx}}$$ of a function defined implicitly by an equation $$F(x,y) = 0$$, provided that $$F$$ is differentiable and $${F_y} \ne 0$$. Prove that if $$F$$ has continuous second derivatives, then a formula for the second derivative of $$y$$ is$$\frac{{{d^2}y}}{{d{x^2}}} = - \frac{{{F_{xx}}F_y^2 - 2{F_{xy}}{F_s}{F_y} + {F_{yy}}F_x^2}}{{F_y^3}}$$

It is proved that the formula for second derivative of $$y$$ is $$\frac{{{d^2}y}}{{d{x^2}}} = - \frac{{{F_y}^2{F_{xx}} + {F_x}^2{F_{yy}} - 2{F_x}{F_y}{F_{xy}}}}{{{F_y}^3}}$$.

See the step by step solution

## Step 1: The Chain Rule and Clairaut’s Theorem

The Chain rule is given by

Suppose that $$z = f(x,y)$$ is a differentiable function of $$x$$ and $$y$$, where $$x = g(s,t)$$ and $$y = h(s,t)$$ are differentiable functions of $$s$$ and $$t$$. then $$\frac{{\partial z}}{{\partial s}} = \frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial s}} + \frac{{\partial z}}{{\partial y}}\frac{{\partial y}}{{\partial s}}$$ and $$\frac{{\partial z}}{{\partial t}} = \frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial t}} + \frac{{\partial z}}{{\partial y}}\frac{{\partial y}}{{\partial t}}$$

Clairaut's Theorem is given by

Suppose $$f$$ is defined on a disk $$D$$ that contains the points $$(a,b)$$. If the function $${f_{xy}}$$ and $${f_{yx}}$$ are both continuous on $$D$$, then $${f_{xy}}(a,b) = {f_{yx}}(a,b)$$.

## Step 2: Use chain rule

The function, $$F(x,y) = 0$$ is differentiable and $${F_y} \ne 0$$.

Equation 6: $$\frac{{dy}}{{dx}} = - \frac{{\frac{{\partial F}}{{dx}}}}{{\frac{{\partial F}}{{\partial y}}}} = - \frac{{{F_x}}}{{{F_y}}}$$.

The implicit function is $$F(x,y) = 0$$.

From equation (6), it is clear that $$\frac{{dy}}{{dx}} = - \frac{{{F_x}}}{{{F_y}}}$$.

Let $$G(x,y) = - \frac{{{F_x}}}{{{F_y}}}$$ then $$\frac{{dy}}{{dx}} = G(x,y)$$.

Thus, $$\frac{{dy}}{{dx}} = G(x,y).(1)$$

Differentiate the equation (1) with respect to $$x$$ on both sides by using chain rule,

$$\frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{d}{{dx}}(G(x,y))$$

$$\frac{{{d^2}y}}{{d{x^2}}} = \frac{{\partial G}}{{\partial x}} \cdot \frac{{dx}}{{dx}} + \frac{{\partial G}}{{\partial y}} \cdot \frac{{dy}}{{dx}}(2)$$

## Step 3: Obtain the value of $$\frac{{\partial G}}{{\partial x}}$$ and $$\frac{{\partial G}}{{\partial y}}$$

Take partial derivative of $$G(x,y) = - \frac{{{F_x}}}{{{F_y}}}$$ with respect to $$x$$ and obtain $$\frac{{\partial G}}{{\partial x}}$$.

\begin{aligned}{l}\frac{{\partial G}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( { - \frac{{{F_x}}}{{{F_y}}}} \right)\\ = - \left( {\frac{{{F_y}{F_{xx}} - {F_x}{F_{yx}}}}{{{{\left( {{F_y}} \right)}^2}}}} \right)\\ = \frac{{{F_x}{F_{yx}} - {F_y}{F_{xx}}}}{{{F_y}^2}}\end{aligned}

Take partial derivative of $$G(x,y) = - \frac{{{F_x}}}{{{F_y}}}$$ with respect to $$y$$ and obtain $$\frac{{\partial G}}{{\partial y}}$$.

\begin{aligned}{l}\frac{{\partial G}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( { - \frac{{{F_x}}}{{{F_y}}}} \right)\\ = - \left( {\frac{{{F_y}{F_{xy}} - {F_x}{F_{yy}}}}{{{{\left( {{F_y}} \right)}^2}}}} \right)\\ = \frac{{{F_x}{F_{yy}} - {F_y}{F_{xy}}}}{{{F_y}^2}}\end{aligned}

## Step 4: Substitute $$\frac{{\partial G}}{{\partial x}}$$ and $$\frac{{\partial G}}{{\partial y}}$$ in the equation (2),

Substitute $$\frac{{\partial G}}{{\partial x}}$$ and $$\frac{{\partial G}}{{\partial y}}$$ in the equation (2) as follows

\begin{aligned}{l}\frac{{{d^2}y}}{{d{x^2}}} = \left( {\frac{{{F_x}{F_{yx}} - {F_y}{F_{xx}}}}{{{F_y}^2}}} \right)\frac{{dx}}{{dx}} + \left( {\frac{{{F_x}{F_{yy}} - {F_y}{F_{xy}}}}{{{F_y}^2}}} \right)\frac{{dy}}{{dx}}\\ = \left( {\frac{{{F_x}{F_{yx}} - {F_y}{F_{xx}}}}{{{F_y}^2}}} \right)(1) + \left( {\frac{{{F_x}{F_{yy}} - {F_y}{F_{xy}}}}{{{F_y}^2}}} \right)\left( { - \frac{{{F_x}}}{{{F_y}}}} \right)\\ = \frac{{{F_x}{F_{yx}} - {F_y}{F_{xx}}}}{{{F_y}^2}} - \frac{{{F_x}^2{F_{yy}}}}{{{F_y}^3}} + \frac{{{F_x}{F_y}{F_{xy}}}}{{{F_y}^3}}\\ = \frac{{{F_x}{F_y}{F_{yx}} - {F_y}^2{F_{xx}} - {F_x}^2{F_{yy}} + {F_x}{F_y}{F_{xy}}}}{{{F_y}^3}}\end{aligned}

## Step 5: Use Clairaut’s theorem

By Clairaut's theorem, it is known that if the function $$F$$ is continuous then $${F_{xy}} = {F_{yx}}$$ are the same.

\begin{aligned}{l}\frac{{{d^2}y}}{{d{x^2}}} = \frac{{{F_x}{F_y}{F_{xy}} - {F_y}^2{F_{xx}} - {F_x}^2{F_{yy}} + {F_x}{F_y}{F_{xy}}}}{{{F_y}^3}}\\ = \frac{{ - {F_y}^2{F_{xx}} - {F_x}^2{F_{yy}} + 2{F_x}{F_y}{F_{xy}}}}{{{F_y}^3}}\\ = - \frac{{{F_y}^2{F_{xx}} + {F_x}^2{F_{yy}} - 2{F_x}{F_y}{F_{xy}}}}{{{F_y}^3}}\end{aligned}

Therefore, it is proved that the formula for second derivative of $$y$$ is $$\frac{{{d^2}y}}{{d{x^2}}} = - \frac{{{F_y}^2{F_{xx}} + {F_x}^2{F_{yy}} - 2{F_x}{F_y}{F_{xy}}}}{{{F_y}^3}}$$.