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Q13E

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Found in: Page 127

Essential Calculus: Early Transcendentals

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

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