Question

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

5.\({x^2} - 4xy + {y^2} = 4\)

Short answer

By Implicit differentiation, \(y' = \frac{{2y - x}}{{y - 2x}}\).

Step-by-step solution

Use Implicit differentiation

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

\(\begin{aligned}\frac{d}{{dx}}\left( {{x^2} - 4xy + {y^2}} \right) &= \frac{d}{{dx}}\left( 4 \right)\\\frac{d}{{dx}}\left( {{x^2}} \right) - \frac{d}{{dx}}\left( {4xy} \right) + \frac{d}{{dx}}\left( {{y^2}} \right) &= \frac{d}{{dx}}\left( 4 \right)\end{aligned}\)

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:

\(\begin{aligned}\frac{d}{{dx}}\left( {{x^2}} \right) - \frac{d}{{dx}}\left( {4xy} \right) + \frac{d}{{dx}}\left( {{y^2}} \right) &= \frac{d}{{dx}}\left( 4 \right)\\2{x^{2 - 1}} - 4\left( {x\frac{{dy}}{{dx}} + y\frac{d}{{dx}}\left( x \right)} \right) + 2y\frac{d}{{dx}}\left( y \right) &= 0\\2x - 4\left( {xy' + y} \right) + 2yy' &= 0\\2x - 4xy' - 4y + 2yy' &= 0\end{aligned}\)

Solve the given equation for \(y'\)

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

\(\begin{aligned}2x - 4xy' - 4y + 2yy'& = 0\\2yy' - 4xy'& = 4y - 2x\\2y'\left( {y - 2x} \right) &= 2\left( {2y - x} \right)\\y'& = \frac{{2y - x}}{{y - 2x}}\end{aligned}\)

Thus, by Implicit differentiation \(y' = \frac{{2y - x}}{{y - 2x}}\).