What is implicit differentiation?

Differentiation can be broadly divided into two categories: explicit differentiation and implicit differentiation. All the differentiation in the AI SL, AI HL and AA SL courses is explicit - we have a function entirely defined in terms of a single variable, and we can differentiate the function explicitly with respect to that variable.

For example, the function $y=x^2 + 2x + 1$, or $f(x) = x^2 + 2x +1$, is a function of $x$, and can be differentiated explicitly with respect to $x$.

$$\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = 2x +2 \,\text{ or }\, f'(x) = 2x + 2 \end{align*}$$

In kinematics, the velocity function $v(t) = \sin (2t)$ is a function of $t$, and can be differentiated explicitly with respect to $t$.

$$\begin{align*} v'(t) &= 2\cos(2t) \end{align*}$$

In the AA HL course, you will encounter implicit differentiation. This arises when we have an equation, or a relationship between two variables, that cannot be expressed as a function of a single variable.

For example, suppose we have the elliptical equation $x^2+y^2-xy=8$. This is not a function, it is a relation, and if we want to find the derivative with respect to $x$, we cannot do so explicitly, because the terms $y^2$ and $-xy$ are not functions of $x$. But we can treat them as functions of $x$ and differentiate them implicitly, using the chain rule (and the product rule).

First consider the term $y^2$. We want to differentiate this with respect to $x$. That is, we want to find $\dfrac{\mathrm{d}}{\mathrm{d}x}\left(y^2\right)$.

If we let $y = f(x)$, so that $\dfrac{\mathrm{d}y}{\mathrm{d}x} = f'(x)$, then what we are trying to find is $\dfrac{\mathrm{d}}{\mathrm{d}x}\left(f(x)\right)^2$.

According to the chain rule, this would be $2f(x)f'(x)$.

Substituting $y$ in place of $f(x)$ and $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ in place of $f'(x)$ we have

$$\begin{align*} \dfrac{\mathrm{d}}{\mathrm{d}x}\left(y^2\right) &= 2f(x)f'(x)\\[8pt] &= 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} \end{align*}$$

We have differentiated $y^2$ with respect to $x$ implicitly.

Now consider the term $-xy$. Again let $y = f(x)$. We want to find $\dfrac{\mathrm{d}}{\mathrm{d}x}\left(-xy\right)$, which we can write as $\dfrac{\mathrm{d}}{\mathrm{d}x}\left(-xf(x)\right)$.

Using the product rule, we have $\dfrac{\mathrm{d}}{\mathrm{d}x}\left(-xf(x)\right) = -f(x) -xf'(x)$.

Substituting $y$ for $f(x)$ and $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ for $f'(x)$ we have found our derivative implicitly.

$$\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\left(-xy\right) &= -f(x) -xf'(x) \\[8pt] &= -y-x\frac{\mathrm{d}y}{\mathrm{d}x} \end{align*}$$

The other two terms, $x^2$ and $8$, can be differentiated explicitly.

$$\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\left(x^2\right) &= 2x \\[8pt] \frac{\mathrm{d}}{\mathrm{d}x} (8) &= 0 \end{align*}$$

So once every term is differentiated we have

$$\begin{align*} 2x + 2y\frac{\mathrm{d}y}{\mathrm{d}x} -y-x\frac{\mathrm{d}y}{\mathrm{d}x} &= 0 \end{align*}$$

Finally, we can make $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ the subject of the equation.

$$\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{2x-y}{y-2x} \end{align*}$$

In general, if we have a function of $y$, but want to differentiate with respect to $x$, we can differentiate it with respect to $y$ then multiply by $\dfrac{\mathrm{d}y}{\mathrm{d}x}$, and that is a reasonable way to understand the implicit differentiation process in the AA HL course.

$$\begin{align*} \dfrac{\mathrm{d}}{\mathrm{d}x}f(y) &= f'(y)\,\dfrac{\mathrm{d}y}{\mathrm{d}x} \end{align*}$$