Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

Answers without the blur.

Just sign up for free and you're in.

Short Answer

Find \({{dy} \mathord{\left/ {\vphantom {{dy} {dx}}} \right.} {dx}}\) by implicit differentiation.
13. \(\sqrt {xy} = 1 + {x^2}y\)

The implicit differentiation of the function \(\sqrt {xy} = 1 + {x^2}y\) is \(\frac{{dy}}{{dx}} = \frac{{4xy\left( {\sqrt {xy} } \right) - y}}{{x - 2{x^2}\left( {\sqrt {xy} } \right)}}\).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Concept of Implicit differentiation and definition of the Chain and product rules.

In Implicit differentiation, first differentiate both sides of the equation with respect to \(x\) and then solve the resulting equation for \(y'\).

If \(y = f\left( u \right)\) and \(u = g\left( x \right)\) are differentiable functions, then the Chain rule is given by \(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}}\)

If \(f\) and \(g\) are differentiable functions, then the product rule is given by \(\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + g\left( x \right)\frac{d}{{dx}}f\left( x \right)\).

Step 2: Find \({{dy} \mathord{\left/ {\vphantom {{dy} {dx}}} \right.} {dx}}\) by implicit differentiation.

The given function is \(\sqrt {xy} = 1 + {x^2}y\).

Differentiate the given function with respect to \(x\) on both sides as follows:

\(\begin{array}{l}\frac{{d\left( {\sqrt {xy} } \right)}}{{dx}} = \frac{{d\left( {1 + {x^2}y} \right)}}{{dx}}\\ \Rightarrow \frac{1}{{2\sqrt {xy} }}\frac{{d\left( {xy} \right)}}{{dx}} = \frac{{d\left( 1 \right)}}{{dx}} + \frac{{d\left( {{x^2}y} \right)}}{{dx}}\end{array}\)

Apply the product rule and the chain rule as follows:

\(\begin{array}{l}\frac{{\rm{1}}}{{{\rm{2}}\sqrt {{\rm{xy}}} }}\left( {\begin{array}{*{20}{c}}{\frac{{{\rm{d}}\left( {\rm{x}} \right)}}{{{\rm{dx}}}}{\rm{y}} + {\rm{x}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}}}\end{array}} \right) = {\rm{0}} + \left( {\frac{{{\rm{d}}\left( {{{\rm{x}}^{\rm{2}}}} \right)}}{{{\rm{dx}}}}{\rm{y}} + {{\rm{x}}^{\rm{2}}}\frac{{{\rm{d}}\left( {\rm{y}} \right)}}{{{\rm{dx}}}}} \right)\\ \Rightarrow \frac{{\rm{1}}}{{{\rm{2}}\sqrt {{\rm{xy}}} }}\left( {{\rm{1y}} + {\rm{x}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right) = {\rm{2}}{{\rm{x}}^{{\rm{2}} - {\rm{1}}}}{\rm{y}} + {{\rm{x}}^{\rm{2}}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}}\\ \Rightarrow \frac{{\rm{y}}}{{{\rm{2}}\sqrt {{\rm{xy}}} }} + \frac{{\rm{x}}}{{{\rm{2}}\sqrt {{\rm{xy}}} }}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} - {{\rm{x}}^{\rm{2}}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = {\rm{2xy}}\\ \Rightarrow \left( {\frac{{\rm{x}}}{{{\rm{2}}\sqrt {{\rm{xy}}} }} - {{\rm{x}}^{\rm{2}}}} \right)\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = {\rm{2xy}} - \frac{{\rm{y}}}{{{\rm{2}}\sqrt {{\rm{xy}}} }}\end{array}\)

Further simplify as follows:

\(\begin{array}{l}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{{\rm{2xy}} - \frac{{\rm{y}}}{{{\rm{2}}\sqrt {{\rm{xy}}} }}}}{{\left( {\frac{{\rm{x}}}{{{\rm{2}}\sqrt {{\rm{xy}}} }} - {{\rm{x}}^{\rm{2}}}} \right)}}\\ \Rightarrow \frac{{dy}}{{dx}} = \frac{{2xy\left( {2\sqrt {xy} } \right) - y}}{{x - {x^2}\left( {2\sqrt {xy} } \right)}}\\ \Rightarrow \frac{{dy}}{{dx}} = \frac{{4xy\left( {\sqrt {xy} } \right) - y}}{{x - 2{x^2}\left( {\sqrt {xy} } \right)}}\end{array}\)

Therefore, the required answer is \(\frac{{dy}}{{dx}} = \frac{{4xy\left( {\sqrt {xy} } \right) - y}}{{x - 2{x^2}\left( {\sqrt {xy} } \right)}}\).

Find Study Materials for

Create Study Materials

Select your language

Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Find \({{dy} \mathord{\left/ {\vphantom {{dy} {dx}}} \right.} {dx}}\) by implicit differentiation.
13. \(\sqrt {xy} = 1 + {x^2}y\)

Step by Step Solution

Step 1: Concept of Implicit differentiation and definition of the Chain and product rules.

Step 2: Find \({{dy} \mathord{\left/ {\vphantom {{dy} {dx}}} \right.} {dx}}\) by implicit differentiation.

Most popular questions for Math Textbooks

Want to see more solutions like these?

Recommended explanations on Math Textbooks

Select your language

Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Find \({{dy} \mathord{\left/ {\vphantom {{dy} {dx}}} \right.} {dx}}\) by implicit differentiation.13. \(\sqrt {xy} = 1 + {x^2}y\)

Step by Step Solution

Step 1: Concept of Implicit differentiation and definition of the Chain and product rules.

Step 2: Find \({{dy} \mathord{\left/ {\vphantom {{dy} {dx}}} \right.} {dx}}\) by implicit differentiation.

Most popular questions for Math Textbooks

Want to see more solutions like these?

Recommended explanations on Math Textbooks

Decision Maths

Calculus

Probability and Statistics

Statistics

Mechanics Maths

Geometry

Pure Maths

Find \({{dy} \mathord{\left/ {\vphantom {{dy} {dx}}} \right.} {dx}}\) by implicit differentiation.
13. \(\sqrt {xy} = 1 + {x^2}y\)