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Essential Calculus: Early Transcendentals
Found in: Page 658
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

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

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Short Answer

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 \(\frac{{{d^2}y}}{{d{x^2}}} = - \frac{{{F_y}^2{F_{xx}} + {F_x}^2{F_{yy}} - 2{F_x}{F_y}{F_{xy}}}}{{{F_y}^3}}\).

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

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 \(\frac{{{d^2}y}}{{d{x^2}}} = - \frac{{{F_y}^2{F_{xx}} + {F_x}^2{F_{yy}} - 2{F_x}{F_y}{F_{xy}}}}{{{F_y}^3}}\).

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