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Implicit Differentiation Basics

We are used to deriving functions with only one variable, usually $x$
However, there are actually more often more than just one variable we need to consider
Differentiating many variables with respect to only one of them is called Implicit Differentiation.
In a sense, all differentiation is implicit; thus, implicit differentiation uses the same rules and techniques that we are familiar with
Remember, $d(anything)$ is an operator; it tells you which variable you are deriving with respect to
If you take the derivative of a variable with respect to itself, the answer is always 1 (i.e. $\frac{dz}{dz}=1$)
$\frac{dy}{dx}\cdot \frac{dx}{dz}=\frac{dy}{dz}$
In most cases, you will be asked to find $dy$; this means you will be deriving $dx$ (with respect to $x$).

Examples of Implicit Differentiation

When faced with a problem involving implicit differentiation, you must take the following course of action:

a) Derive both sides of the equation
b) use algebra to separate and solve for the desired derivative

Suppose you have an equation $2x^3+x^2y+y^2=4$, and you are asked to find the derivative of $y$. This implies that you will be deriving with respect to $x$.
First step: derive the entire function using the chain and product rules: $2x^3+x^2y+y^2=4 \rightarrow 6x^2+2xy+y'x^2+2yy'=0$.
Now, use algebra to separate what we want to solve for by pulling out the common factor ($y'$): $6x^2+2xy+y'(x^2+2y)=0$
Finally, just solve the equation: $y'=\frac{-6x^2-2xy}{x^2+2y}$. This is the equation for $\frac{dy}{dx}$.

Parameters of Geometric Figures

\begin{array}{|c|c|c|} \hline \textrm{Shape} & \textrm{Volume} & \textrm{Surface Area}\\ \hline \textrm{Circle} & \textrm{Area} =\pi r^2 & \textrm{circumference} = 2\pi r \\ \hline \textrm{Cylinder} & V=\pi r^2 h & SA=2\pi r h +2 \pi r^2\\ \hline \textrm{Cone} & V=\frac{1}{3}\pi r^2 h & SA=\pi r l + \pi r^2 \\ \hline \textrm{Sphere} & V=\frac{4}{3} \pi r^3 & SA=4 \pi r^2 \\ \hline \textrm{Cube} & V=a^3 & SA=6a^2\\ \hline \textrm{Pyramid} & V= \frac{1}{3} lwh & SA=\frac{1}{2}Pl+B \\ \hline \end{array}

Related Rates

Related rates are applications of derivatives; thus, they often deal with real-world situations.
If you are faced with a related rates problem, this generally means that you are dealing with two or more things that are changing at the same time and are affecting each other as they do so.
An example would be a faucet filling up a conical well. The water is leaving the faucet at a constant rate, while the well is filling slower as it gets more full.
Here are the steps to solving problems involving related rates:
Identify that it is a related rates problems: In general, problems dealing with related rates will include some type of geometrical figure, whether 2D or 3D. However, this does not have to be the case. A related rates problem is any problem where you are dealing with two or more variables changing at the same time.
Write down your known and unknown variables: this only takes a second, and can be immensely helpful, especially when dealing with more variables.
Write function: write a function in which relates all of the variables.
Derive: Take the derivative of the function. Since there is more than one variable, you will be using implicit differentiation. Most of the time, you will be deriving with respect to time, $dt$ (since real world rates are measured like this, i.e. kilometers per hour).
Substitute: plug your known variables back into the derivative equation.
Solve: use algebra to solve for the variable or rate that is needed.


Here is an example of a basic related rates problem:
A spherical balloon is losing volume at a rate of $\frac{3}{\pi} \frac{m^3}{s}$. At the moment the volume of the balloon becomes $\frac{256}{3\pi^2}m^3$, at what rate is the radius changing in $\frac{m}{s}$?
Here is the solution:
First step, identify the problem. Since I already told you that this is a related rates problem, consider step one done
Write out knowns and unknowns: $\frac{dV}{dt}=\frac{3}{\pi}$; $V=\frac{256}{3\pi^2}$; $\frac{dr}{dt}=?$.
Write the function you will be using. In this case, this is the volume function of a sphere, $V=\frac{4}{3}\pi r^3$.
Derive this function with implicit differentiation: $\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$.
Substitute: At this point, it becomes evident to us that we do not have enough information to solve the problem (we don't know the radius $r$). However, we can simply find it using the equation for volume that was given to us ($V=\frac{256}{3\pi^2}$). Setting this equal to $\frac{4}{3}\pi r^3$ (the volume formula), we get that $r=\frac{4}{\pi}$. Now, our list of knowns reads: $\frac{dV}{dt}=\frac{3}{\pi}$; $V=\frac{256}{3\pi^2}$; $r=\frac{4}{\pi}$; and we have all the information necessary to solve the equation.
Solve: Substituting our knowns back into the equation, we get $-\frac{3}{\pi}=4\pi \cdot\frac{16}{\pi^2}\frac{dr}{dt}$ ($\frac{dV}{dt}$ is negative because the balloon is losing volume). Now solving, we get that $\frac{dr}{dt}=-\frac{3}{64}$