Question

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

8.\({x^3} - x{y^2} + {y^3} = 1\)

Short answer

By Implicit differentiation, \(y' = \frac{{{y^2} - 3{x^2}}}{{y\left( {3y - 2x} \right)}}\).

Step-by-step solution

Use Implicit differentiation

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

\(\begin{aligned}\frac{d}{{dx}}\left( {{x^3} - x{y^2} + {y^3}} \right) &= \frac{d}{{dx}}\left( 1 \right)\\\frac{d}{{dx}}\left( {{x^3}} \right) - \frac{d}{{dx}}\left( {x{y^2}} \right) + \frac{d}{{dx}}\left( {{y^3}} \right) &= \frac{d}{{dx}}\left( 1 \right)\end{aligned}\)

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:

\(\begin{aligned}\frac{d}{{dx}}\left( {{x^3}} \right) - \frac{d}{{dx}}\left( {x{y^2}} \right) + \frac{d}{{dx}}\left( {{y^3}} \right) &= \frac{d}{{dx}}\left( 1 \right)\\3{x^2} - \left( {2xyy' + {y^2}} \right) + 3{y^2}y' &= 0\\3{x^2} - 2xyy' - {y^2} + 3{y^2}y' &= 0\end{aligned}\)

Solve the given equation for \(y'\) 

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

\(\begin{aligned}{c}3{x^2} - 2xyy' - {y^2} + 3{y^2}y' &= 0\\y'\left( {3{y^2} - 2xy} \right) &= {y^2} - 3{x^2}\\y' &= \frac{{{y^2} - 3{x^2}}}{{3{y^2} - 2xy}}\\y' &= \frac{{{y^2} - 3{x^2}}}{{y\left( {3y - 2x} \right)}}\end{aligned}\)

Thus, by Implicit differentiation\(y' = \frac{{{y^2} - 3{x^2}}}{{y\left( {3y - 2x} \right)}}\).