Question

Use implicit differentiation to find\(dy/dx\)for the equation

\(\frac{x}{y} = {y^2} + 1\;\;\;\;\;\;\;\;\;y \ne 0\)

and for the equivalent equation

\(x = {y^3} + y\;\;\;\;\;\;\;\;y \ne 0\)

Show that although the expressions you get for\(dy/dx\)look different, they agree for all points that satisfy the given equation.

Short answer

It is proved that the expressions you get for \(\frac{{dy}}{{dx}}\) look different. They agree for all points that satisfy the given equation.

Step-by-step solution

Use Implicit differentiation

As \(y\) is the function of \(x\), differentiating both sides of the equation \(\frac{x}{y} = {y^2} + 1\) with respect to \(x\).

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

Therefore, by using the implicit method, \(y' = \frac{y}{{2{y^3} + x}}\) for the equation \(\frac{x}{y} = {y^2} + 1\).

Use Implicit differentiation

As \(y\) is the function of \(x\), differentiate both sides of the equation \(x = {y^3} + y\) with respect to \(x\).

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

Therefore, by using the implicit method, \(y' = \frac{1}{{3{y^2} + 1}}\) for the equation \(x = {y^3} + y\).

Show that both the obtained expression of \(y'\) satisfying the same points

Substitute\(x = {y^3} + y\) into the equation \(y' = \frac{y}{{2{y^3} + x}}\) and simplify.

\(\begin{aligned}{c}y' &= \frac{y}{{2{y^3} + {y^3} + y}}\\ &= \frac{y}{{3{y^3} + y}}\\ &= \frac{y}{{y\left( {3{y^2} + 1} \right)}}\\ &= \frac{1}{{3{y^2} + 1}}\end{aligned}\)

Thus, both expression satisfies the same points.