What is Implicit Differentiation?
In many calculus problems, you’ll see equations that don’t exactly represent functions. Normally, we’re used to seeing equations like $$y = 2x + 1$$ where $y$ is isolated on one side. However, what if \(x\) and \(y\) are mixed together, like $$x^2 + y^2 = 1 { or } 3x^2y -4x\cos y = 3e^{xy}$$ and it is impossible to solve for $y$ in terms of $x$? This type of equation is called an implicit equation. In this case, to find $\dfrac {dy}{dx}$ we use implicit differentiation.
An Example
Suppose we want to find $\dfrac {dy}{dx}$ given $x^2 + y^2 = 1$ with respect to $x$. We have:
\[\begin{align*} \dfrac{d}{dx}(x^2 + y^2) &= \dfrac{d}{dx}(1) \\ \dfrac{d}{dx}(x^2) + \dfrac{d}{dx}(y^2) &= \dfrac{d}{dx}(1) \end{align*}\]
To begin, we can take the derivative of $x^2$ and $1$ by using the power rule, and we would get $2x$ and $0$ respectively:
\[\begin{align*} 2x + \dfrac{d}{dx}(y^2) &= 0 \end{align*}\]
But how do we find the derivative $\dfrac{d}{dx}(y^2)$? A common mistake is to apply the power rule and arrive at $$\dfrac{d}{dx}(y^2) = 2y$$ However, this is \emph{not} true! It would be correct if we derived $y$ with respect to $y$:
\[\begin{align*} \dfrac{d}{dx}(y^2) \not= 2y \:\:\:\:\:\:\: \dfrac{d}{dy}(y^2) = 2y \end{align*}\]
To implicitly differentiate $y$, we must pretend that \emph{$y$ is a function of $x$}. Here are some examples of functions of $x$:
$$y = x + 1\qquad\sqrt{x}\qquad-x^2 – 3$$ This is counterintuitive, since we cannot solve for $y$ in terms of $x$. This is what makes implicit differentiation so powerful. To emphasize that $y$ is a function of $x$, we will write it as $y(x)$ just for this example.
To find $\dfrac{d}{dx}(y^2)$ we use the chain rule! Recall that for functions $f(x)$ and $g(x)$ the chain rule states that
\[\begin{align*} \dfrac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x) \end{align*}\]
We let $f(x) = x^2$ and $g(x) = y(x)$. Using the power rule, $f'(x) = 2x$, so
\[\begin{align*} \dfrac{d}{dx}(y(x)^2) = \dfrac{d}{dx} (f(y(x))) = f'(y(x)) \cdot y'(x) = 2y(x) \cdot y'(x) \end{align*}\]
Putting it all together, implicitly differentiating $x^2 + y^2 = 1$ gets us:
\[\begin{align*} \dfrac{d}{dx}(x^2 + y^2) &= \dfrac{d}{dx}(1) \\ 2x + 2y &= 0\\ 2y &=-2x\\ y &=\boxed{-\frac xy} \end{align*}\]
Explanation of Implicit Differentiation
Implicit differentiation is really not that hard. You just need to treat $y$ as a function of $x$. Remember to use the chain rule and product rule appropriately. As an example, let’s differentiate both sides of
\[\begin{align*} x\tan x = x^2e^{\tan x} \end{align*}\]
We will use the chain rule and product rule. The derivative of the left side is
\[\begin{align*} \tan x + x\sec^2 x \end{align*}\]
and the derivative of the right side is
\[\begin{align*} 2xe^{\tan x} + x^2 e^{\tan x} \sec^2 x \end{align*}\]
Combining the two yields
\[\begin{align*} \cos x + x\sin x = 2xe^{\tan x} + x^2 e^{\tan x} \sec^2 x \end{align*}\]
Let us now compare this to differentiating implicitly
\[\begin{align*} xy = x^2e^y \end{align*}\]
The answer would be
\[\begin{align*} y + x\dfrac{dy}{dx} = 2x e^y + x^2 e^y \dfrac{dy}{dx} \end{align*}\]
Put more simply, when differentiating $y$ with respect to $x$, treat $y$ as $x$, but then add $\dfrac{dy}{dx}$ at the end.
Another Example
Suppose we want to implicitly differentiate $$x^2 + y^3 = x + 1$$
If you were to just differentiate $y^3$, you would get $3y^2$. Remember, however, that $y$ is a function of $x$ and you need to add $\dfrac{dy}{dx}$ at the end. Therefore, if we carefully go through the steps, the process would be
\[\begin{align*} \dfrac{d}{dx}(x^2 + y^3) &= \dfrac{d}{dx}(x + 1) \\ \dfrac{d}{dx}(x^2) + \dfrac{d}{dx}(y^3) &= \dfrac{d}{dx}(x) + \dfrac{d}{dx}(1) \\ 2x + 3y^2 \dfrac{dy}{dx} &= 1 + 0\\ 3y^2 \dfrac{dy}{dx} &= 1-2x\\ \dfrac{dy}{dx} &= \boxed{\frac{1-2x}{3y^2}} \end{align*}\]
