Question

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

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

Short answer

By Implicit differentiation, \(y' = \frac{{1 - {e^y}}}{{x{e^y} + 1}}\)

Step-by-step solution

Use Implicit differentiation 

Differentiate both sides of the equation with respect to \(x\).

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

Product rule of differentiation 

The Product rule of differentiation implies that \(\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}\left( {g\left( x \right)} \right) + g\left( x \right)\frac{d}{{dx}}\left( {f\left( x \right)} \right)\), where \(f\) and \(g\) are differentiable and use formula \(\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\).

As \(y\) is the function of \(x\), use product rule and simplify as:

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

Solve the given equation for

Simplify the obtained equation for \(y'\).

\(\begin{aligned}x{e^y}y' + y' &= 1 - {e^y}\\\left( {x{e^y} + 1} \right)y' &= 1 - {e^y}\\y' &= \frac{{1 - {e^y}}}{{x{e^y} + 1}}\end{aligned}\)

Thus, by Implicit differentiation \(y' = \frac{{1 - {e^y}}}{{x{e^y} + 1}}\).