Implicit Differentiation (DP IB Analysis & Approaches (AA): HL): Revision Note

Implicit differentiation

What is implicit differentiation?

• It is not always easy or possible to write an equation connecting x and y explicitly in the form $y=f\left(x\right)$ or $x=f\left(y\right)$

  • In such cases the equation is written implicitly

    • as a function of $x$ and $y$

    • e.g. in the form $f(x,y)=0$

• The parts of such equations involving $y$ can be differentiated implicitly using the chain rule

$\frac{\mathrm{d}}{dx}\left[f\left(y\right)\right]=f^{\text{'}}\left(y\right)\frac{\mathrm{d}y}{\mathrm{d}x}$

• A shortcut way of thinking about this is that ‘$y$ is (implicitly) a function of $x$’

  • And when differentiating a function of $y$ chain rule says “differentiate with respect to $y$, then multiply by the derivative of $y$” (which is $\frac{\mathrm{d}y}{\mathrm{d}x}$)

How do I find a derivative using implicit differentiation?

• In order to find $\frac{\mathrm{d}y}{\mathrm{d}x}$ for an implicit relation between $x$ and $y$, follow these steps

• STEP 1
  Differentiate both sides of the equation implicitly with respect to $x$
  Use $\frac{\mathrm{d}}{dx}\left[f\left(y\right)\right]=f^{\text{'}}\left(y\right)\frac{\mathrm{d}y}{\mathrm{d}x}$ for terms containing $y$

  • E.g.  $2x^2+y^4=48$

$\begin{array}{rcl}\frac{\mathrm{d}}{\mathrm{d}x}\left(2x^2+y^4\right)& =& \frac{\mathrm{d}}{\mathrm{d}x}\left(48\right)\\ 4x+4y^3\frac{\mathrm{d}y}{\mathrm{d}x}& =& 0\end{array}$
 

• STEP 2
  Rearrange the new equation to make $\frac{\mathrm{d}y}{\mathrm{d}x}$ the subject

$\begin{array}{rcl}\frac{\mathrm{d}y}{\mathrm{d}x}& =& -\frac{x}{y^3}\end{array}$

Applications of implicit differentiation

What type of problems could involve implicit differentiation?

• Broadly speaking there are three types of problem that could involve implicit differentiation

  • Algebraic problems involving graphs, derivatives, tangents, normals, etc.,

    where it is not practical to write $y$ explicitly in terms of $x$

    • Usually in such cases, $\frac{\mathrm{d}y}{\mathrm{d}x}$ will be in terms of $x$ and $y$

  • Optimisation problems that involve time derivatives

    • More than one variable may be involved too

      e.g.  Volume of a cylinder, $V=\pi r^{2}h$

      e.g.  The side length and (therefore) area of a square increase over time

  • Any problem that involves differentiating with respect to an extraneous variable

    • e.g. $y=f\left(x\right)$ but the derivative $\frac{\mathrm{d}y}{\mathrm{d}\theta }$ is required (rather than $\frac{\mathrm{d}y}{\mathrm{d}x}$)

How do I apply implicit differentiation to algebraic problems?

• Algebraic problems revolve around values of the derivative (gradient) $\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)$

• Particular problems focus on special case tangent values

  • Horizontal tangents

    • Also referred to as tangents parallel to the $x$-axis

    • This is when $\frac{\mathrm{d}y}{\mathrm{d}x}=0$

  • Vertical tangents

    • Also referred to as tangents parallel to the $y$-axis

    • This is when $\frac{\mathrm{d}x}{\mathrm{d}y}=0$

  • Recall the relation $\frac{\mathrm{d}x}{\mathrm{d}y}=\frac{1}{{\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}}}$

• Other problems may involve finding equations of (other) tangents and/or normals

• For problems that involve finding the coordinates of points on a curve with a specified gradient the method below can be used
   

• STEP 1
  Differentiate the equation of the curve implicitly
   

• STEP 2
  Substitute the given or implied value of $\frac{\mathrm{d}y}{\mathrm{d}x}$ to create an equation linking $x$ and $y$
   

• STEP 3
  There are now two equations

  • the original equation

  • the linking equation

  Solve them simultaneously to find the $x$ and $y$ coordinates as required

How do I apply implicit differentiation to optimisation problems?

• For a single variable use chain rule to differentiate implicitly

  • e.g. A square with side length changing over time, $A=x^2$

    • Differentiating both sides with respect to $t$ gives

$\frac{\mathrm{d}A}{\mathrm{d}t}=2x\frac{\mathrm{d}x}{\mathrm{d}t}$

• For more than one variable use product rule (and chain rule) to differentiate implicitly

  • e.g. A square-based pyramid with base length $x$ and height $h$ changing over time, $V=\frac{1}{3}{x}^{2}h$

    • Differentiating both sides with respect to $t$ gives

$\frac{\mathrm{d}V}{\mathrm{d}t}=\frac{1}{3}\left[x^2\frac{\mathrm{d}h}{\mathrm{d}t}+2x\frac{\mathrm{d}x}{\mathrm{d}t}h\right]=\frac{1}{3}x\left(x\frac{\mathrm{d}h}{\mathrm{d}t}+2h\frac{\mathrm{d}x}{\mathrm{d}t}\right)$

• After differentiating implicitly the rest of the question should be similar to any other optimisation problem

  • Be aware of phrasing

    • “the rate of change of the height of the pyramid” (over time) is $\frac{\mathrm{d}h}{\mathrm{d}t}$

  • When finding the location of minimum and maximum problems

    • The solution is not necessarily at a turning point

    • The minimum or maximum could be at the start or end of a given or appropriate interval