Figure 2.5.3(a)   Folium of Descartes and its dependence on the parameter $a$ 

Solution by Context Panel

$x^3\+y^3\=3 a x y$$\stackrel{\text{implicit differentiation}}{\to }$$\frac{ay-x^2}{-ax\+y^2}$

Interactively implemented stepwise solution

$x^3\+y^3\=3 a x y$

$x^3\+y^3\=3axy$

$\stackrel{\text{evaluate at point}}{\to }$

$x^3\+{y\left(x\right)}^3\=3axy\left(x\right)$

$\stackrel{\text{differentiate w.r.t. x}}{\to }$

$3x^2\+3{y\left(x\right)}^2\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)\=3ay\left(x\right)\+3ax\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)$

$\stackrel{\text{isolate for diff(y(x),x)}}{\to }$

$\frac{\ⅆ}{\ⅆx}y\left(x\right)\=\frac{-3x^2\+3ay\left(x\right)}{3{y\left(x\right)}^2-3ax}$

Solution via the implicitdiff command

$\mathrm{implicitdiff}\left(x^3\+y^3\=3 a x y\,y\,x\right)$ = $\frac{ay-x^2}{-ax\+y^2}$$$

Stepwise solution via the ImplicitDiffSolution command

$\mathrm{Student}:-\mathrm{Calculus1}:-\mathrm{ImplicitDiffSolution}\left(x^3\+y^3\=3 a x y\,y \,x\right)$

$\begin{array}{lll}& & \text{Implicit Differentiation Steps}\\ & & x^3+y^3=3axy\\ \text{•}& & \text{Rewrite}y\text{as a function}y\left(x\right)\text{:}\\ & & x^3+{y\left(x\right)}^3=3axy\left(x\right)\\ \text{•}& & \text{Differentiate the left side}\\ & & \frac{\ⅆ}{\ⅆx}\left(x^3+{y\left(x\right)}^3\right)\\ \text{▫}& & \text{1. Apply the}\mathbf{\text{sum}}\text{rule}\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{sum}}\text{rule}\\ & & \frac{\ⅆ}{\ⅆx}\left(f\left(x\right)+g\left(x\right)\right)=\frac{\ⅆ}{\ⅆx}f\left(x\right)+\frac{\ⅆ}{\ⅆx}g\left(x\right)\\ & & f\left(x\right)=x^3\\ & & g\left(x\right)={y\left(x\right)}^3\\ & & \text{This gives:}\\ & & \frac{\ⅆ}{\ⅆx}\left(x^3\right)+\frac{\ⅆ}{\ⅆx}\left({y\left(x\right)}^3\right)\\ \text{▫}& & \text{2. Apply the}\mathbf{\text{power}}\text{rule to the term}\frac{\ⅆ}{\ⅆx}\left(x^3\right)\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{power}}\text{rule}\\ & & \frac{\partial }{\partial x}\left(x^n\right)=n{x}^{n-1}\\ & \text{◦}& \text{This means:}\\ & & \frac{\ⅆ}{\ⅆx}\left(x^3\right)=3\cdot x^2\\ & & \text{We can rewrite the derivative as:}\\ & & \left(3x^2\right)+\frac{\ⅆ}{\ⅆx}\left({y\left(x\right)}^3\right)\\ \text{▫}& & \text{3. Apply the}\mathbf{\text{chain}}\text{rule to the term}{y\left(x\right)}^3\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{chain}}\text{rule}\\ & & \frac{\ⅆ}{\ⅆx}f\left(g\left(x\right)\right)=\mathrm{f\text{'}}\left(g\left(x\right)\right)\left(\frac{\ⅆ}{\ⅆx}g\left(x\right)\right)\\ & \text{◦}& \text{Outside function}\\ & & f\left(v\right)=v^3\\ & \text{◦}& \text{Inside function}\\ & & g\left(x\right)=y\left(x\right)\\ & \text{◦}& \text{Derivative of outside function}\\ & & \frac{\ⅆ}{\ⅆv}f\left(v\right)=3v^2\\ & \text{◦}& \text{Apply composition}\\ & & \mathrm{f\text{'}}\left(g\left(x\right)\right)=3{y\left(x\right)}^2\\ & \text{◦}& \text{Derivative of inside function}\\ & & \frac{\ⅆ}{\ⅆx}g\left(x\right)=\frac{\ⅆ}{\ⅆx}y\left(x\right)\\ & \text{◦}& \text{Put it all together}\\ & & \left(\frac{\ⅆ}{\ⅆx}f\left(g\left(x\right)\right)\right)\left(\frac{\ⅆ}{\ⅆx}g\left(x\right)\right)=3{y\left(x\right)}^2\cdot \left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)\\ & & \text{This gives:}\\ & & 3x^2+\left(3{y\left(x\right)}^2\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)\right)\\ \text{•}& & \text{The final result is}\\ & & 3x^2+3\cdot {y\left(x\right)}^2\cdot \left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)\\ \text{•}& & \text{Differentiate the right side}\\ & & \frac{\partial }{\partial x}\left(3axy\left(x\right)\right)\\ \text{▫}& & \text{1. Apply the}\mathbf{\text{constant multiple}}\text{rule to the term}\frac{\partial }{\partial x}\left(3axy\left(x\right)\right)\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant multiple}}\text{rule}\\ & & \frac{\partial }{\partial x}\left(Cf\left(x\right)\right)=C\left(\frac{\ⅆ}{\ⅆx}f\left(x\right)\right)\\ & \text{◦}& \text{This means:}\\ & & \frac{\partial }{\partial x}\left(3axy\left(x\right)\right)=3a\cdot \left(\frac{\ⅆ}{\ⅆx}\left(xy\left(x\right)\right)\right)\\ & & \text{We can rewrite the derivative as:}\\ & & 3a\left(\frac{\ⅆ}{\ⅆx}\left(xy\left(x\right)\right)\right)\\ \text{▫}& & \text{2. Apply the}\mathbf{\text{product}}\text{rule}\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{product}}\text{rule}\\ & & \frac{\ⅆ}{\ⅆx}\left(f\left(x\right)g\left(x\right)\right)=\left(\frac{\ⅆ}{\ⅆx}f\left(x\right)\right)g\left(x\right)+f\left(x\right)\left(\frac{\ⅆ}{\ⅆx}g\left(x\right)\right)\\ & & f\left(x\right)=x\\ & & g\left(x\right)=y\left(x\right)\\ & & \text{This gives:}\\ & & 3a\left(\left(\frac{\ⅆx}{\ⅆx}\right)y\left(x\right)+x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)\right)\\ \text{▫}& & \text{3. Apply the}\mathbf{\text{power}}\text{rule to the term}\frac{\ⅆx}{\ⅆx}\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{power}}\text{rule}\\ & & \frac{\partial }{\partial x}\left(x^n\right)=n{x}^{n-1}\\ & \text{◦}& \text{This means:}\\ & & \frac{\ⅆ}{\ⅆx}{x}^1=1\cdot x^0\\ & \text{◦}& \text{So,}\\ & & \frac{\ⅆx}{\ⅆx}=1\\ & & \text{We can rewrite the derivative as:}\\ & & 3\cdot a\cdot \left(\left(y\left(x\right)\right)+x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)\right)\\ \text{•}& & \text{The final result is}\\ & & 3\cdot a\cdot \left(y\left(x\right)+x\cdot \left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)\right)\\ \text{•}& & \text{Rewrite}\frac{\ⅆ}{\ⅆx}y\left(x\right)\text{as}\mathrm{y\text{'}}\text{and solve for}\mathrm{y\text{'}}\\ & & 3x^2+3\cdot y^2\cdot \mathrm{y\text{'}}=3\cdot a\cdot \left(y+x\cdot \mathrm{y\text{'}}\right)\\ \text{•}& & \text{Subtract}3\cdot a\cdot \left(y+x\cdot \mathrm{y\text{'}}\right)\text{from both sides}\\ & & 3\cdot x^2+3\cdot y^2\cdot \mathrm{y\text{'}}-3\cdot a\cdot \left(y+x\cdot \mathrm{y\text{'}}\right)=3\cdot a\cdot \left(y+x\cdot \mathrm{y\text{'}}\right)-3\cdot a\cdot \left(y+x\cdot \mathrm{y\text{'}}\right)\\ \text{•}& & \text{Simplify}\\ & & 3\cdot x^2+3\cdot y^2\cdot \mathrm{y\text{'}}-3\cdot a\cdot \left(y+x\cdot \mathrm{y\text{'}}\right)=0\\ \text{•}& & \text{Subtract}3\cdot x^2\text{from both sides}\\ & & 3\cdot x^2+3\cdot y^2\cdot \mathrm{y\text{'}}-1\cdot \left(3\cdot a\cdot \left(y+x\cdot \mathrm{y\text{'}}\right)\right)-3\cdot x^2=0-3\cdot x^2\\ \text{•}& & \text{Simplify}\\ & & 3\cdot y^2\cdot \mathrm{y\text{'}}-1\cdot \left(3\cdot a\cdot \left(y+x\cdot \mathrm{y\text{'}}\right)\right)=\left(\mathrm{-3}\right)\cdot x^2\\ \text{•}& & \text{Reorder terms}\\ & & 3y^2\mathrm{y\text{'}}-1\cdot \left(3\cdot a\cdot \left(x\mathrm{y\text{'}}+y\right)\right)=\left(\mathrm{-3}\right)\cdot x^2\\ \text{•}& & \text{Distributive multiply}\\ & & 3y^2\mathrm{y\text{'}}-1\cdot \left(3\cdot a\cdot \left(x\mathrm{y\text{'}}\right)+3\cdot a\cdot y\right)=\left(\mathrm{-3}\right)\cdot x^2\\ \text{•}& & \text{Distributive multiply}\\ & & 3y^2\mathrm{y\text{'}}+\left(\left(\mathrm{-1}\right)\cdot \left(3ax\mathrm{y\text{'}}\right)-1\cdot \left(3ay\right)\right)=\left(\mathrm{-3}\right)\cdot x^2\\ \text{•}& & \text{Multiply constants}\\ & & 3y^2\mathrm{y\text{'}}+\left(\left(\mathrm{-3}\right)\cdot a\cdot x\cdot \mathrm{y\text{'}}-1\cdot \left(3ay\right)\right)=\left(\mathrm{-3}\right)\cdot x^2\\ \text{•}& & \text{Multiply constants}\\ & & 3y^2\mathrm{y\text{'}}+\left(-3ax\mathrm{y\text{'}}-3\cdot a\cdot y\right)=\left(\mathrm{-3}\right)\cdot x^2\\ \text{•}& & \text{Reorder terms}\\ & & -3ax\mathrm{y\text{'}}+3y^2\mathrm{y\text{'}}-3ay=\left(\mathrm{-3}\right)\cdot x^2\\ \text{•}& & \text{Add}3ay\text{to both sides}\\ & & -3ax\mathrm{y\text{'}}+3y^2\mathrm{y\text{'}}-3ay+3ay=\left(\mathrm{-3}\right)\cdot x^2+3ay\\ \text{•}& & \text{Simplify}\\ & & -3ax\mathrm{y\text{'}}+3y^2\mathrm{y\text{'}}=\left(\mathrm{-3}\right)\cdot x^2+3ay\\ \text{•}& & \text{Reorder terms}\\ & & -3ax\mathrm{y\text{'}}+3y^2\mathrm{y\text{'}}=3ay-3x^2\\ \text{•}& & \text{Factor}\\ & & \mathrm{y\text{'}}\cdot \left(-3ax+3y^2\right)=3ay-3x^2\\ \text{•}& & \text{Divide both sides by}-3ax+3y^2\\ & & \frac{\mathrm{y\text{'}}\cdot \left(-3ax+3y^2\right)}{-3ax+3y^2}=\frac{3ay-3x^2}{-3ax+3y^2}\\ \text{•}& & \text{Simplify}\\ & & \mathrm{y\text{'}}=\frac{3ay-3x^2}{-3ax+3y^2}\\ \text{•}& & \text{Solution}\\ & & \mathrm{y\text{'}}=-\frac{ay-x^2}{ax-y^2}\end{array}$

$x^3\+y^3\=3 a x y$

$\stackrel{\text{implicit differentiation}}{\to }$

$\frac{ay-x^2}{-ax\+y^2}$

$\stackrel{\text{numerator}}{\to }$

$\,\=0$

$x^3\+y^3\=3axy\,-ay\+x^2\=0$

$\stackrel{\text{solve (specified)}}{\to }$

$\left\{x\=0\,y\=0\right\}\,\left\{x\=\mathrm{RootOf}\left({\mathrm{\_Z}}^3-2\right)a\,y\={\mathrm{RootOf}\left({\mathrm{\_Z}}^3-2\right)}^{2}a\right\}$

$\stackrel{\text{radical form}}{\=}$

$\left\{\left\{x\=0\,y\=0\right\}\,\left\{x\=2^{1\/3}a\,y\=2^{2\/3}a\right\}\right\}$

Figure 2.5.3(a)   Folium and horizontal tangents with $a\=-1$

Figure 2.5.3(b)   Folium and horizontal tangents with $a\=1$

$x^3\+y^3\=3 a x y$

$\stackrel{\text{implicit differentiation}}{\to }$

$\frac{ay-x^2}{-ax\+y^2}$

$\stackrel{\text{denominator}}{\to }$

$\,\=0$

$x^3\+y^3\=3axy\,ax-y^2\=0$

$\stackrel{\text{solve (specified)}}{\to }$

$\left\{x\=0\,y\=0\right\}\,\left\{x\={\mathrm{RootOf}\left({\mathrm{\_Z}}^3-2\right)}^{2}a\,y\=\mathrm{RootOf}\left({\mathrm{\_Z}}^3-2\right)a\right\}$

$\stackrel{\text{radical form}}{\=}$

$\left\{\left\{x\=0\,y\=0\right\}\,\left\{x\=2^{2\/3}a\,y\=2^{1\/3}a\right\}\right\}$

Figure 2.5.3(c)   Folium and vertical tangents with $a\=-1$

Figure 2.5.3(d)   Folium and vertical tangents with $a\=1$

$x^3\+y^3\=9 x y$

$\stackrel{\text{implicit differentiation}}{\to }$

$\,\=1$

$x^3\+y^3\=9xy\,\frac{-x^2\+3y}{y^2-3x}\=1$

$\stackrel{\text{evaluate at point}}{\to }$

$\stackrel{\text{solve}}{\to }$

$u^3\+v^3\=9uv\,\frac{-u^2\+3v}{v^2-3u}\=1$

$\stackrel{\text{solve}}{\to }$

$y\=1\cdot \left(x-u\right)\+v$

$\genfrac{}{}{0ex}{}{}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{}$

$y\=x-2.043750116$

$\genfrac{}{}{0ex}{}{}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{}$

$y\=x\+2.043750116$

$A\={\left(-4 x^3\+4\sqrt{-108x^3\+x^6}\right)}^{1\/3}$ 

$x^3\+y^3\=9 x y$

$x^3\+y^3\=9xy$

$\stackrel{\text{solutions for y}}{\to }$

$\frac{1}{2}{\left(-4x^3\+4\sqrt{x^6-108x^3}\right)}^{1\/3}\+\frac{6x}{{\left(-4x^3\+4\sqrt{x^6-108x^3}\right)}^{1\/3}}\,-\frac{1}{4}{\left(-4x^3\+4\sqrt{x^6-108x^3}\right)}^{1\/3}-\frac{3x}{{\left(-4x^3\+4\sqrt{x^6-108x^3}\right)}^{1\/3}}\+\frac{1}{2}\mathrm{I}\sqrt{3}\left(\frac{1}{2}{\left(-4x^3\+4\sqrt{x^6-108x^3}\right)}^{1\/3}-\frac{6x}{{\left(-4x^3\+4\sqrt{x^6-108x^3}\right)}^{1\/3}}\right)\,-\frac{1}{4}{\left(-4x^3\+4\sqrt{x^6-108x^3}\right)}^{1\/3}-\frac{3x}{{\left(-4x^3\+4\sqrt{x^6-108x^3}\right)}^{1\/3}}-\frac{1}{2}\mathrm{I}\sqrt{3}\left(\frac{1}{2}{\left(-4x^3\+4\sqrt{x^6-108x^3}\right)}^{1\/3}-\frac{6x}{{\left(-4x^3\+4\sqrt{x^6-108x^3}\right)}^{1\/3}}\right)$

Table 2.5.3(b)   Interactive steps for obtaining Figure 2.5.3(f)