Question

5–22: Find \(dy/dx\) by implicit differentiation.

21. \({e^{\frac{x}{y}}} = x - y\)

Short answer

By Implicit differentiation, \(\frac{{dy}}{{dx}} = \frac{{y\left( {y - {e^{\frac{x}{y}}}} \right)}}{{{y^2} - x{e^{\frac{x}{y}}}}}\).

Step-by-step solution

Write the definition of implicit differentiation and the formula of the quotient rule

Implicit differentiation: Differentiate the both sides of the equation with respect to \(x\) and then solve the obtained result for \(\frac{{dy}}{{dx}}\) this process is known asImplicit differentiation.

The Quotient rule: If \(f\) and \(g\) are differentiable functions, then \(\frac{d}{{dx}}\left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \frac{{g\left( x \right)\frac{d}{{dx}}\left( {f\left( x \right)} \right) + f\left( x \right)\frac{d}{{dx}}\left( {g\left( x \right)} \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\).

Find \(\frac{{dy}}{{dx}}\)by implicit differentiation

Consider the equation, \({e^{\frac{x}{y}}} = x - y\). Differentiate both the sides of the equation w.r.t. \(x\) and simplify by using quotient rule.

\(\begin{aligned}\frac{d}{{dx}}\left( {{e^{\frac{x}{y}}}} \right) &= \frac{d}{{dx}}\left( {x - y} \right)\\{e^{\frac{x}{y}}}\frac{d}{{dx}}\left( {\frac{x}{y}} \right) &= 1 - \frac{{dy}}{{dx}}\\{e^{\frac{x}{y}}}\left( {\frac{{y\left( 1 \right) - x\frac{{dy}}{{dx}}}}{{{y^2}}}} \right) &= 1 - \frac{{dy}}{{dx}}\\{e^{\frac{x}{y}}} \cdot \left( {\frac{1}{y}} \right) - \frac{{x{e^{\frac{x}{y}}}}}{{{y^2}}} \cdot \frac{{dy}}{{dx}} &= 1 - \frac{{dy}}{{dx}}\\\frac{{dy}}{{dx}} - \frac{{x{e^{\frac{x}{y}}}}}{{{y^2}}} \cdot \frac{{dy}}{{dx}} &= 1 - {e^{\frac{x}{y}}} \cdot \left( {\frac{1}{y}} \right)\end{aligned}\)

Furthermore,

\(\begin{aligned}\frac{{dy}}{{dx}}\left( {1 - \frac{{x{e^{\frac{x}{y}}}}}{{{y^2}}}} \right) &= \frac{{y - {e^{\frac{x}{y}}}}}{y}\\\frac{{dy}}{{dx}} &= \frac{{\frac{{y - {e^{\frac{x}{y}}}}}{y}}}{{1 - \frac{{x{e^{\frac{x}{y}}}}}{{{y^2}}}}}\\\frac{{dy}}{{dx}} &= \frac{{y\left( {y - {e^{\frac{x}{y}}}} \right)}}{{{y^2} - x{e^{\frac{x}{y}}}}}\end{aligned}\)

Thus, \(\frac{{dy}}{{dx}} = \frac{{y\left( {y - {e^{\frac{x}{y}}}} \right)}}{{{y^2} - x{e^{\frac{x}{y}}}}}\).