Question

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

7.\({x^4} + {x^2}{y^2} + {y^3} = 5\)

Short answer

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

Step-by-step solution

Use Implicit differentiation

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

\(\begin{aligned}\frac{d}{{dx}}\left( {{x^4} + {x^2}{y^2} + {y^3}} \right) &= \frac{d}{{dx}}\left( 5 \right)\\\frac{d}{{dx}}\left( {{x^4}} \right) + \frac{d}{{dx}}\left( {{x^2}{y^2}} \right) + \frac{d}{{dx}}\left( {{y^3}} \right) &= \frac{d}{{dx}}\left( 5 \right)\end{aligned}\)

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:

\(\begin{aligned}\frac{d}{{dx}}\left( {{x^4}} \right) + \frac{d}{{dx}}\left( {{x^2}{y^2}} \right) + \frac{d}{{dx}}\left( {{y^3}} \right) &= \frac{d}{{dx}}\left( 5 \right)\\4{x^{4 - 1}} + \left( {{x^2}\frac{d}{{dx}}\left( {{y^2}} \right) + {y^2}\frac{d}{{dx}}\left( {{x^2}} \right)} \right) + 3{y^{3 - 1}}\frac{d}{{dx}}\left( y \right) &= 0\\4{x^3} + 2{x^2}yy' + 2x{y^2} + 3{y^2}y' &= 0\end{aligned}\)

Solve the given equation for \(y'\)

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

\(\begin{aligned}4{x^3} + 2{x^2}yy' + 2x{y^2} + 3{y^2}y' &= 0\\\left( {2{x^2}y + 3{y^2}} \right)y'& = - \left( {4{x^3} + 2x{y^2}} \right)\\y' &= - \frac{{\left( {4{x^3} + 2x{y^2}} \right)}}{{\left( {2{x^2}y + 3{y^2}} \right)}}\\y'& = - \frac{{2x\left( {2{x^2} + {y^2}} \right)}}{{y\left( {2{x^2} + 3y} \right)}}\end{aligned}\)

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