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Implicit Differentation

Now that we’ve seen some examples of taking derivatives of function, we will look at a new situation. Here, we will be given an equation that defines a relation between different variables and asked to find what the derivative of one with respect to another would be. Since the relation is defined implicitly, we call this implicit differentiation.

Implicit Differentiation

Here we will need to find the derivative of \(y\) with respect to \(x\) when

In the past, we have been given \(y\) as a function of \(x\), so we could just take the derivative using the rules that we had accumulated in class. In this situation; however, we will not be able to find \(y\) as a function of \(x\) explicitly. Instead, what we will do is treat \(y\) as if it is some unknown function of \(x\), let’s say \(y(x)\). Then, since both sides of the equation are equal, we will have that if we take the derivative of both sides with respect to \(x\), the equality will be maintained. We then have

As we look at this, there is a lot going on. Instead of trying to do everything at once, we will break this up into smaller portions. If we begin by distributing the derivative across sums, we find that we will need the derivatives of

  • \(y^{3}\),
  • \(x^{3}\),
  • \(xe^{y}\).

We will now look at each of these pieces.

Derivative of \(y^{3}\)

As we look at \(y^{3}\) we need to remember that \(y\) is a function of \(x\). Therefore, if we want to find \(\frac{d}{dx}y^{3}\), we must look at this as

Looking at this example, we begin by looking at the big picture. Here, there is indeed a function inside of a function, namely \(y(x)\) is inside \(i^{3}\), so we will need to use the chain rule. To do this, we find

  • \(\frac{do}{dx}=\frac{do}{di}\frac{di}{dx}\),
  • \(o(i)=i^{3}\),
  • \(i(x)=y(x)\),
  • \(\frac{do}{di}=3i^{2}\),
  • \(\frac{di}{dx}=\frac{dy}{dx}\).

Note that for the derivative of the inside, we don’t know what \(y\) is. Therefore, the bests we can do is say that the derivative of the inside is just the derivative of \(y\) with respect to \(x\). Combining these together and recalling that \(i=y\), we find that

Derivative of \(x^{3}\)

Here we note that we are finding \(\frac{d}{dx}x^{3}\). Since the variable in the equation, (\(x\)), are the same as the variable we are taking the derivative with respect to, we need only take the normal derivative here. That is

Derivative of \(xe^{y}\)

In order to start this derivative, we must first note that we are taking the derivative of a product of two functions. We, therefore, start with the product rule and get

  • \((FS)’=SF’+FS’\)
  • \(F=x\)
  • \(S=e^{y}\)
  • \(F’=1\)
  • \(S’=?\)

Note that, since \(F\) was in terms of \(x\), we could take the normal derivative of \(x\). However, since \(S\) is in terms of \(y\), we need to do more work. We will, therefore, break this off as a separate problem.

Derivative of \(e^{y}\)

As we look at \(e^{y}\), we note, as we did with \(y^{3}\) that \(y\) is a function of \(x\), say \(y(x)\). Because of this, \(e^{y(x)}\) is indeed a function inside a function. We will then use the chain rule to get

  • \(\frac{do}{dx}=\frac{do}{di}\frac{di}{dx}\),
  • \(o(i)=e^{i}\),
  • \(i(x)=y\),
  • \(\frac{do}{di}=e^{i}\),
  • \(\frac{di}{dx}=\frac{dy}{dx}\).

Combining these, we see that

Finishing \(xe^{y}\)

Now that we have found the derivative of \(e^{y}\), we can go back to the problem of finding the derivative of \(xe^{y}\). As we do this, we get
\frac{d}{dx}xe^{y}&=SF’+FS’ \\

Finding \(\frac{dy}{dx}\)

We have now found the derivative of each of the pieces, so we are ready to find \(\frac{dy}{dx}\). We now see that
\frac{d}{dx}(y^{3}+x^{3})&=\frac{d}{dx}xe^{y} \\

We now group the terms with \(\frac{dy}{dx}\) on one side, and everything else on the other. After this, we factor out the \(\frac{dy}{dx}\) and divide by the coefficient. This gives us
\frac{d}{dx}(y^{3}+x^{3})&=\frac{d}{dx}xe^{y} \\
3y^{2}\frac{dy}{dx}-xe^{y}\frac{dy}{dx}&=e^{y}-3x^{2} \\
\frac{dy}{dx}(3y^{2}-xe^{y})&=e^{y}-3x^{2} \\

We have now found the derivative of \(y\) with respect to \(x\). I do want to point out that, while we did solve for \(\frac{dy}{dx}\), we have given this in terms of both \(x\) and \(y\). Therefore, if we want to find the derivative at a given point, we would need to know both the \(x\) and \(y\) values.


In this problem, we saw that we could find \(\frac{dy}{dx}\) even if we can’t explicitly write \(y\) as a function of \(x\). We did so by knowing that if two functions are equal (or two sides of an equation) the derivative of both sides must also be equal. This allowed us to find terms involving \(\frac{dy}{dx}\) so that we could solve in terms of both \(x\) and \(y\).

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