Let’s consider, as an example, the function $e^{x^2}$. When you take its derivative, you of course use the Chain Rule:

\begin{align*}

\dfrac{d}{dx}\left(e^{x^2} \right) &= e^{x^2} \cdot \dfrac{d}{dx}\left(x^2 \right) \\[8px]&= e^{x^2} \cdot (2x)

\end{align*}

Similarly, when you take the derivative of, say, $e^{f(x)}$, where $f(x)$ is some function we’re not specifying, you again of course use the Chain Rule:

\begin{align*}

\dfrac{d}{dx}\left(e^{f(x)} \right) &= e^{f(x)} \cdot \dfrac{d}{dx}\left(f(x) \right) \\[8px]&= e^{f(x)} \cdot \dfrac{df}{dx}

\end{align*}

That might look funny, but we don’t know what $\dfrac{df}{dx}$ is, so we’re just leaving it written like that.

And also similarly, let’s consider the derivative of, say, $e^{y(x)}$, where we’re writing $y(x)$ to indicate that *y *is a function of *x*. We again of course must use the Chain Rule:

\begin{align*}

\dfrac{d}{dx}\left(e^{y(x)} \right) &= e^{y(x)} \cdot \dfrac{d}{dx}\left(y(x) \right) \\[8px]&= e^{y(x)} \cdot \dfrac{dy(x)}{dx}

\end{align*}

Now we don’t usually write $y(x)$, because it’s annoying to write again and again; instead we just write *“y”*—but always have to keep in mind that really *y *still depends on *x*. We thus usually write that last result as

$$\dfrac{d}{dx}\left(e^y \right) = e^y \cdot \dfrac{dy}{dx}$$

The key point is that we end up multiplying by $\dfrac{dy}{dx}$, because the function we were taking the derivative of, $e^y$, has a *y *in it. Similarly, you should see that the following are all correct derivatives for these example functions:

\begin{align*}

\dfrac{d}{dx}\left(y \right) &= 1 \cdot \dfrac{dy}{dx} = \dfrac{dy}{dx}\\[8px]\dfrac{d}{dx}\left(y^5 \right) &= 5y^4 \cdot \dfrac{dy}{dx} \\[8px]\dfrac{d}{dx}\left( \sin(y) \right) &= \cos(y) \cdot \dfrac{dy}{dx} \\[8px]\dfrac{d}{dx}\left( \cos(y^2) \right) &= -\sin(y^2) \cdot (2y) \cdot \dfrac{dy}{dx} \\[8px]\dfrac{d}{dx}\left(\ln(y) \right) &= \frac{1}{y} \cdot \dfrac{dy}{dx} \\[8px]\dfrac{d}{dx}\left(\ln(y^5) \right) &= \frac{1}{y^5} \cdot \left(5y^4 \right) \cdot \dfrac{dy}{dx} = \frac{5}{y} \cdot\dfrac{dy}{dx}

\end{align*}

The biggest challenge when learning to do Implicit Differentiation problems is to remember to include this $\dfrac{dy}{dx}$ term when you take the derivative of something that has a *y* in it. As always, practicing is the way to learn, and you’ll get good practice problems below.

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