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Essential Calculus: Early Transcendentals
Found in: Page 127
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

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

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 \(\frac{{dy}}{{dx}} = \frac{{4xy\left( {\sqrt {xy} } \right) - y}}{{x - 2{x^2}\left( {\sqrt {xy} } \right)}}\).

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