The ImplicitDiffSolution command computes the partial derivative of the function, y with respect to x, showing the steps required to make the computation. The input f defines y as a function of x implicitly. It must be an equation in x and y or an algebraic expression, which is understood to be equated to zero.

All other names, which appear in the input f and the derivative variable(s) x and are not of type constant, are treated as independent variables.

Optional arguments output, displaystyle, and animated can be passed to control the style of output.  These options are described in Student:-Basics:-OutputStepsRecord. The return value is controlled by the output option.

This function is part of the Student:-Calculus1 package.

$\mathrm{with}\left(\mathrm{Student}:-\mathrm{Calculus1}\right)\:$

$\mathrm{ImplicitDiffSolution}\left(x^2+y^3=1\,y\,x\right)$

$\begin{array}{lll}& & \text{Implicit Differentiation Steps}\\ & & y^3+x^2=1\\ \text{•}& & \text{Rewrite}y\text{as a function}y\left(x\right)\text{:}\\ & & {y\left(x\right)}^3+x^2=1\\ \text{•}& & \text{Differentiate the left side}\\ & & \frac{\ⅆ}{\ⅆx}\left({y\left(x\right)}^3+x^2\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)={y\left(x\right)}^3\\ & & g\left(x\right)=x^2\\ & & \text{This gives:}\\ & & \frac{\ⅆ}{\ⅆx}\left({y\left(x\right)}^3\right)+\frac{\ⅆ}{\ⅆx}\left(x^2\right)\\ \text{▫}& & \text{2. Apply the}\mathbf{\text{power}}\text{rule to the term}\frac{\ⅆ}{\ⅆx}\left(x^2\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^2\right)=2\cdot x^1\\ & \text{◦}& \text{So,}\\ & & \frac{\ⅆ}{\ⅆx}\left(x^2\right)=2\cdot x\\ & & \text{We can rewrite the derivative as:}\\ & & \frac{\ⅆ}{\ⅆx}\left({y\left(x\right)}^3\right)+\left(2x\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:}\\ & & \left(3{y\left(x\right)}^2\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)\right)+2x\\ \text{•}& & \text{The final result is}\\ & & 3\cdot {y\left(x\right)}^2\cdot \left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+2x\\ \text{•}& & \text{Differentiate the right side}\\ & & \frac{\ⅆ}{\ⅆx}1\\ \text{▫}& & \text{4. Apply the}\mathbf{\text{constant}}\text{rule to the term}\frac{\ⅆ}{\ⅆx}1\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant}}\text{rule}\\ & & \frac{\ⅆC}{\ⅆx}=0\\ & \text{◦}& \text{This means:}\\ & & \frac{\ⅆ}{\ⅆx}1=0\\ & & \text{We can now rewrite the derivative as:}\\ & & 0\\ \text{•}& & \text{Rewrite}\frac{\ⅆ}{\ⅆx}y\left(x\right)\text{as}\mathrm{y\text{'}}\text{and solve for}\mathrm{y\text{'}}\\ & & 3\cdot y^2\cdot \mathrm{y\text{'}}+2x=0\\ \text{•}& & \text{Subtract}2\cdot x\text{from both sides}\\ & & 3\cdot y^2\cdot \mathrm{y\text{'}}+2\cdot x-2\cdot x=0-2\cdot x\\ \text{•}& & \text{Simplify}\\ & & 3\cdot y^2\cdot \mathrm{y\text{'}}=\left(\mathrm{-2}\right)\cdot x\\ \text{•}& & \text{Divide both sides by}3\cdot y^2\\ & & \frac{\mathrm{y\text{'}}\cdot \left(3\cdot y^2\right)}{3\cdot y^2}=\frac{\left(\mathrm{-2}\right)\cdot x}{3\cdot y^2}\\ \text{•}& & \text{Simplify}\\ & & \mathrm{y\text{'}}=\frac{-2x}{3\cdot y^2}\\ \text{•}& & \text{Solution}\\ & & \mathrm{y\text{'}}=-\frac{2x}{3y^2}\end{array}$

$\mathrm{ImplicitDiffSolution}\left(a{x}^{3}y-\frac{2y}{z}=z^2\,y\left(x\,z\right)\,x\right)$

$\begin{array}{lll}& & \text{Implicit Differentiation Steps}\\ & & a{x}^{3}y-\frac{2y}{z}=z^2\\ \text{•}& & \text{Rewrite}y\text{as a function}y\left(x\,z\right)\text{:}\\ & & a{x}^{3}y\left(x\,z\right)-\frac{2y\left(x\,z\right)}{z}=z^2\\ \text{•}& & \text{Differentiate the left side}\\ & & \frac{\partial }{\partial x}\left(a{x}^{3}y\left(x\,z\right)-\frac{2y\left(x\,z\right)}{z}\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)=a{x}^{3}y\left(x\,z\right)\\ & & g\left(x\right)=-\frac{2y\left(x\,z\right)}{z}\\ & & \text{This gives:}\\ & & \frac{\partial }{\partial x}\left(a{x}^{3}y\left(x\,z\right)\right)+\frac{\partial }{\partial x}\left(-\frac{2y\left(x\,z\right)}{z}\right)\\ \text{▫}& & \text{2. Apply the}\mathbf{\text{constant multiple}}\text{rule to the term}\frac{\partial }{\partial x}\left(a{x}^{3}y\left(x\,z\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(a{x}^{3}y\left(x\,z\right)\right)=a\cdot \left(\frac{\partial }{\partial x}\left(x^{3}y\left(x\,z\right)\right)\right)\\ & & \text{We can rewrite the derivative as:}\\ & & \left(a\left(\frac{\partial }{\partial x}\left(x^{3}y\left(x\,z\right)\right)\right)\right)+\frac{\partial }{\partial x}\left(-\frac{2y\left(x\,z\right)}{z}\right)\\ \text{▫}& & \text{3. 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^3\\ & & g\left(x\right)=y\left(x\,z\right)\\ & & \text{This gives:}\\ & & a\left(\left(\frac{\ⅆ}{\ⅆx}\left(x^3\right)\right)y\left(x\,z\right)+x^3\left(\frac{\partial }{\partial x}y\left(x\,z\right)\right)\right)+\frac{\partial }{\partial x}\left(-\frac{2y\left(x\,z\right)}{z}\right)\\ \text{▫}& & \text{4. 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:}\\ & & a\cdot \left(\left(3\cdot x^2\right)\cdot y\left(x\,z\right)+x^3\left(\frac{\partial }{\partial x}y\left(x\,z\right)\right)\right)+\frac{\partial }{\partial x}\left(-\frac{2y\left(x\,z\right)}{z}\right)\\ \text{▫}& & \text{5. Apply the}\mathbf{\text{constant multiple}}\text{rule to the term}\frac{\partial }{\partial x}\left(-\frac{2y\left(x\,z\right)}{z}\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(-\frac{2y\left(x\,z\right)}{z}\right)=-\frac{2}{z}\cdot \left(\frac{\partial }{\partial x}y\left(x\,z\right)\right)\\ & & \text{We can rewrite the derivative as:}\\ & & a\left(3x^{2}y\left(x\,z\right)+x^3\left(\frac{\partial }{\partial x}y\left(x\,z\right)\right)\right)+\left(-\frac{2\left(\frac{\partial }{\partial x}y\left(x\,z\right)\right)}{z}\right)\\ \text{•}& & \text{The final result is}\\ & & a\left(3x^{2}y\left(x\,z\right)+x^3\left(\frac{\partial }{\partial x}y\left(x\,z\right)\right)\right)-2\cdot \frac{1}{z}\cdot \left(\frac{\partial }{\partial x}y\left(x\,z\right)\right)\\ \text{•}& & \text{Differentiate the right side}\\ & & \frac{\partial }{\partial x}\left(z^2\right)\\ \text{▫}& & \text{6. Apply the}\mathbf{\text{constant}}\text{rule to the term}\frac{\partial }{\partial x}\left(z^2\right)\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant}}\text{rule}\\ & & \frac{\ⅆC}{\ⅆx}=0\\ & \text{◦}& \text{This means:}\\ & & \frac{\partial }{\partial x}\left(z^2\right)=0\\ & & \text{We can now rewrite the derivative as:}\\ & & 0\\ \text{•}& & \text{Rewrite}\frac{\partial }{\partial x}y\left(x\,z\right)\text{as}\mathrm{y\text{'}}\text{and solve for}\mathrm{y\text{'}}\\ & & a\left(x^3\mathrm{y\text{'}}+3x^{2}y\right)-2\cdot \frac{1}{z}\cdot \mathrm{y\text{'}}=0\\ \text{•}& & \text{Multiply through:}a\cdot \left(x^3\mathrm{y\text{'}}+3x^{2}y\right)\text{=}a{x}^3\mathrm{y\text{'}}+3a{x}^{2}y\\ & & a{x}^3\mathrm{y\text{'}}+3a{x}^{2}y+\frac{\left(\mathrm{-2}\right)\cdot \mathrm{y\text{'}}}{z}=0\\ \text{•}& & \text{Subtract}3a{x}^{2}y\text{from both sides}\\ & & a{x}^3\mathrm{y\text{'}}+3a{x}^{2}y+\frac{\left(\mathrm{-2}\right)\cdot \mathrm{y\text{'}}}{z}-3a{x}^{2}y=0-3a{x}^{2}y\\ \text{•}& & \text{Simplify}\\ & & a{x}^3\mathrm{y\text{'}}+\frac{\left(\mathrm{-2}\right)\cdot \mathrm{y\text{'}}}{z}=-3a{x}^{2}y\\ \text{•}& & \text{Find common denominator}\\ & & \frac{z\cdot \left(a{x}^3\mathrm{y\text{'}}\right)}{z}+\frac{\left(\mathrm{-2}\right)\cdot \mathrm{y\text{'}}}{z}=-3a{x}^{2}y\\ \text{•}& & \text{Sum over common denominator}\\ & & \frac{z\cdot \left(a{x}^3\mathrm{y\text{'}}\right)-2\cdot \mathrm{y\text{'}}}{z}=-3a{x}^{2}y\\ \text{•}& & \text{Multiply rhs by denominator of lhs}\\ & & \frac{z\cdot \left(a{x}^3\mathrm{y\text{'}}\right)-2\cdot \mathrm{y\text{'}}}{z}\cdot z=z\cdot \left(-3a{x}^{2}y\right)\\ \text{•}& & \text{Simplify}\\ & & z\cdot \left(a{x}^3\mathrm{y\text{'}}\right)-2\cdot \mathrm{y\text{'}}=-3a{x}^{2}yz\\ \text{•}& & \text{Factor}\\ & & \mathrm{y\text{'}}\cdot \left(a{x}^{3}z-2\right)=-3a{x}^{2}yz\\ \text{•}& & \text{Divide both sides by}a{x}^{3}z-2\\ & & \frac{\mathrm{y\text{'}}\cdot \left(a{x}^{3}z-2\right)}{a{x}^{3}z-2}=\frac{-3a{x}^{2}yz}{a{x}^{3}z-2}\\ \text{•}& & \text{Simplify}\\ & & \mathrm{y\text{'}}=\frac{-3a{x}^{2}yz}{a{x}^{3}z-2}\\ \text{•}& & \text{Solution}\\ & & \mathrm{y\text{'}}=-\frac{3a{x}^{2}yz}{a{x}^{3}z-2}\end{array}$

$\mathrm{Understand}\left(\mathrm{diff}\,\mathrm{constant}\,\mathrm{power}\,\mathrm{constantmultiple}\right)$

$\mathrm{Diff}=\left[\mathrm{constant}\,\mathrm{power}\,\mathrm{constantmultiple}\right]$

$\mathrm{ImplicitDiffSolution}\left(y^3+x^2=1\,y\,x\right)$

$\begin{array}{lll}& & \text{Implicit Differentiation Steps}\\ & & y^3+x^2=1\\ \text{•}& & \text{Rewrite}y\text{as a function}y\left(x\right)\text{:}\\ & & {y\left(x\right)}^3+x^2=1\\ \text{•}& & \text{Differentiate the left side}\\ & & \frac{\ⅆ}{\ⅆx}\left({y\left(x\right)}^3+x^2\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)={y\left(x\right)}^3\\ & & g\left(x\right)=x^2\\ & & \text{This gives:}\\ & & \frac{\ⅆ}{\ⅆx}\left({y\left(x\right)}^3\right)+\frac{\ⅆ}{\ⅆx}\left(x^2\right)\\ \text{•}& & \text{2. Apply the}\mathbf{\text{power}}\text{rule to the term}\frac{\ⅆ}{\ⅆx}\left(x^2\right)\\ & & \frac{\ⅆ}{\ⅆx}\left({y\left(x\right)}^3\right)+\left(2x\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:}\\ & & \left(3{y\left(x\right)}^2\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)\right)+2x\\ \text{•}& & \text{The final result is}\\ & & 3\cdot {y\left(x\right)}^2\cdot \left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+2x\\ \text{•}& & \text{Differentiate the right side}\\ & & \frac{\ⅆ}{\ⅆx}1\\ \text{•}& & \text{4. Apply the}\mathbf{\text{constant}}\text{rule to the term}\frac{\ⅆ}{\ⅆx}1\\ & & 0\\ \text{•}& & \text{Rewrite}\frac{\ⅆ}{\ⅆx}y\left(x\right)\text{as}\mathrm{y\text{'}}\text{and solve for}\mathrm{y\text{'}}\\ & & 3\cdot y^2\cdot \mathrm{y\text{'}}+2x=0\\ \text{•}& & \text{Subtract}2\cdot x\text{from both sides}\\ & & 3\cdot y^2\cdot \mathrm{y\text{'}}+2\cdot x-2\cdot x=0-2\cdot x\\ \text{•}& & \text{Simplify}\\ & & 3\cdot y^2\cdot \mathrm{y\text{'}}=\left(\mathrm{-2}\right)\cdot x\\ \text{•}& & \text{Divide both sides by}3\cdot y^2\\ & & \frac{\mathrm{y\text{'}}\cdot \left(3\cdot y^2\right)}{3\cdot y^2}=\frac{\left(\mathrm{-2}\right)\cdot x}{3\cdot y^2}\\ \text{•}& & \text{Simplify}\\ & & \mathrm{y\text{'}}=\frac{-2x}{3\cdot y^2}\\ \text{•}& & \text{Solution}\\ & & \mathrm{y\text{'}}=-\frac{2x}{3y^2}\end{array}$

The Student:-Calculus1:-ImplicitDiffSolution command was introduced in Maple 2023.

For more information on Maple 2023 changes, see Updates in Maple 2023.