# Implicit Differentiation – Explicitly Explained

Up to this point, you should have a fair understanding of how to find derivatives of functions. Usually, these functions are written as $$f(x)$$ and you’ll find $$f'(x)$$, or equivalently, the function may be called $$y$$ and you’ll find $$y’$$, where $$f(x)$$ or $$y$$ is always written in terms of $$x$$. Sometimes, though, you may need to take the derivative of a function that is not in this format. For example, the relation $$2xy-y^{2}=5x$$ features multiple $$y’s$$, and cannot be rearranged to the regular “$$y=$$“ format. Even so, using a process called “implicit differentiation,” we can find the derivative $$\frac{dy}{dx}$$.

As I mentioned before, many of the functions that require implicit differentiation are not functions at all because they often cannot be expressed in the regular “$$y=$$” format and consequently do not pass the vertical line test. For this reason, it is more appropriate to call such equations “relations” instead of “functions.”

Let’s start by considering the relation $$6xy-x^{2}=3y$$ and finding $$\frac{dy}{dx}$$ through the process of implicit differentiation.

To do this, we will want to take the derivative of each term in the relation. We can start by writing “$$\frac{d}{dx}$$” in front of every single term. Remember, $$\frac{d}{dx}$$ is just a notation for saying “take the derivative with respect to $$x$$.”

STEP 1: Write $$\frac{d}{dx}$$ in front of all terms.

When we do this, we get $$\frac{d}{dx}6xy-\frac{d}{dx}x^{2}=\frac{d}{dx}3y$$.

The second step in implicit differentiation is taking each of these derivatives.

STEP 2: Take the derivative of each term.

Let’s look at these derivatives one at a time. First, we have the derivative of $$6xy$$ with respect to $$x$$. How can we take the derivative with respect to $$x$$ if we have a $$y$$, though? To do this, let’s remember that when we talk about $$y$$, we typically are talking about $$y$$ itself being a function of $$x$$. With implicit differentiation, the same idea applies. We are going to consider all $$y$$’s to be implicit functions of $$x$$. For this term then, we have $$6x$$ times a function of $$x$$. As with derivatives of other products, this calls for the product rule. “First times the derivative of the second” gives us $$6x$$ times $$\frac{dy}{dx}$$, because the derivative of $$y$$, $$\frac{dd}{xy}$$, can be written as $$\frac{dy}{dx}$$. Continuing with the product rule, we add “second times the derivative of the first,” which gives us $$y$$ times 6. Putting these together, we have $$6x\frac{dy}{dx}+6y$$ as the derivative of the first term.

The next derivative is $$-\frac{d}{dx}x^{2}$$. Since this term doesn’t have any $$y$$’s, we can just take its derivative normally and get $$-2x$$.

On the right side, we have the derivative $$\frac{d}{dx}3y$$. Remember, since the derivative of y is $$\frac{dy}{dx}$$, we get $$3\frac{dy}{dx}$$.

Putting all of the pieces together, we now have $$6x\frac{dy}{dx}+6y-2x=3\frac{dy}{dx}$$. Time for the third and final step of implicit differentiation. Since the goal is finding $$\frac{dy}{dx}$$, we now just need to isolate $$\frac{dy}{dx}$$ on one side of the equation.

STEP 3: Solve for $$\frac{dy}{dx}$$.

Let’s move $$6x\frac{dy}{dx}$$ to the other side of the equation so we can have all the $$\frac{dy}{dx}$$ together. To do this, we subtract $$6x\frac{dy}{dx}$$ from both sides of the equation.

$$6y-2x=3\frac{dy}{dx}-6x\frac{dy}{dx}$$

From here, we can factor out the $$\frac{dy}{dx}$$.

$$6y-2x=\frac{dy}{dx}(3-6x)$$

Then, divide both sides by $$3-6x$$.

$$\frac{6y-2x}{3-6x}=\frac{dy}{dx}$$

And just like that we have the solution! The three steps of implicit differentiation are writing $$\frac{d}{dx}$$ in front of each term, taking the derivative of each term with respect to $$x$$, and then solving for $$\frac{dy}{dx}$$.

Let’s try this with another example. Use implicit differentiation to find $$\frac{dy}{dx}$$ for the relation $$y^3-4x=xy$$. Just like in the last example, we can start the implicit differentiation process by writing $$\frac{d}{dx}$$ in front of each term. This gives us:

$$\frac{d}{dx}y^3-\frac{d}{dx}4x=\frac{d}{dx}xy$$

Next, we need to take the derivative of each term with respect to $$x$$. Notice that the first term is $$y^3$$. This is different from $$y^1$$, which we could take the derivative of just by writing $$\frac{dy}{dx}$$. Because we are taking the derivative with respect to $$x$$, we have to be careful and remember that $$y$$ is a function of $$x$$. In the first term, we are essentially seeing a function, $$y$$, inside another function, a cube. Sound familiar?

Because we have a function inside a function, we need to use the chain rule for this term!

Remember, the chain rule says that the derivative of a composite function $$f(g(x))$$ is equal to the derivative of the outside function with the inside function unchanged, times the derivative of the inside function.

$$(f(g(x)))’=f'(g(x))\cdot g'(x)$$

So in this problem, the derivative of $$y^3$$ with respect to $$x$$ is equal to the derivative of the outer cube function, $$3y^2$$, times the derivative of the inner function $$y$$.

$$3y^2\cdot \frac{dy}{dx}$$

This is a very important step to remember with implicit differentiation since it comes up quite frequently. In fact, any time you have to take the derivative of a term involving y to some power other than 1, you’ll need to use the chain rule and multiply by $$\frac{dy}{dx}$$.

Let’s take the derivatives of the other terms now. The derivative of -4x with respect to $$x$$ is of course -4. On the right side, the derivative of $$xy$$ with respect to $$x$$ can be found using the product rule.

“First times the derivative of the second” gives us $$\frac{dy}{dx}$$, “plus second times the derivative of the first,” which is $$y1$$. So we get $$x\cdot \frac{dy}{dx}+y$$. Putting everything together, we now have:

$$3y^2\frac{dy}{dx}-4=x\frac{dy}{dx}+y$$

The final step of implicit differentiation is solving this equation for $$\frac{dy}{dx}$$. To do this, let’s move $$x\frac{dy}{dx}$$ to the left side, and move the -4 to the right side. By subtracting $$x\frac{dy}{dx}$$ from both sides and adding 4 to both sides, we get this:

$$3y^2\frac{dy}{dx}-x\frac{dy}{dx}=y+4$$

Now, let’s factor out $$\frac{dy}{dx}$$ from both terms on the left side.

$$\frac{dy}{dx}(3y^2-x)=y+4$$

Finally, we can get $$\frac{dy}{dx}$$ by itself by dividing both sides by $$3y^2-x$$.

$$\frac{dy}{dx}=\frac{y+4}{3y^2-x}$$

You may have noticed that both of the solutions in this lesson so far have included the letter $$y$$ in the derivative. This is nothing to be worried about, though, and is just a consequence of how implicit differentiation works. Rest assured, as long as you follow the three steps properly and remember the chain rule where applicable, you’ll be on the right track.

Let’s work through one more example. Use implicit differentiation to find $$\frac{dy}{dx}$$ for the relation $$\text{sin}(y^2)=5x^2+y$$.

First, we can start by writing $$\frac{d}{dx}$$ in front of each term.

$$\frac{d}{dx}\text{sin}(y^2)=\frac{d}{dx}5×2+\frac{d}{dx}y$$

The first term is going to be the trickiest. Because we have a function, $$y^2$$, inside the sine function, we are going to need the chain rule. The derivative of the outside will be cosine, so we are going to have $$cos(^y2)$$ times the derivative of the inside.

$$\text{cos}(y^2)\cdot (\frac{d}{dx}y^2)$$

But what is the derivative of $$y^2$$ with respect to $$x$$? Since $$y$$ is not to the first power, we are actually going to need to use the chain rule a second time. Now, the outer function is the square, while the inner function is $$y$$. So the derivative of $$y^2$$ with respect to $$x$$ will be $$2y$$ times $$\frac{dy}{dx}$$.

$$\frac{d}{dx}y^2=2y\frac{dy}{dx}$$

We can now rearrange this a little bit on the left side and we’ll get $$2y \hspace{2 pt}\text{cos}(y^2)\frac{dy}{dx}$$ as the derivative of the first term.

Fortunately, the other two derivatives in this problem are much simpler. We of course know that the derivative $$5x^2$$ is $$10x$$, and that the derivative $$y$$ can just be written as $$\frac{dy}{dx}$$. Overall, we now have:

$$2y \hspace{2 pt}\text{cos}(y^2)\frac{dy}{dx}=10x+\frac{dy}{dx}$$

The final step is, once again, to solve for $$\frac{dy}{dx}$$. To do this, let’s subtract $$\frac{dy}{dx}$$ from the right side to the left side.

$$2y \hspace{2 pt}\text{cos}(y^2)\frac{dy}{dx}-\frac{dy}{dx}=10x$$

Then we can factor out $$\frac{dy}{dx}$$ from the terms on the left side to get:

$$\frac{dy}{dx}(2y \hspace{2 pt}\text{cos}(y^2)-1)=10x$$

Dividing both sides by $$2y \hspace{2 pt}\text{cos}(y^2)-1$$, we get the solution:

$$\frac{dy}{dx}=\frac{10x}2y \hspace{2 pt}\text{cos}(y^2)-1$$

Implicit differentiation can seem like a weird and tricky topic in calculus, but keep in mind the three main steps. First, write $$\frac{d}{dx}$$ in front of all terms in the relation. Second, take the derivative of each term, taking special care to remember that $$y$$ is a function of $$x$$ and will often require the chain rule for proper execution. Third, solve the resulting equation for $$\frac{dy}{dx}$$. As always, with a little practice, you should feel more and more confident with implicit differentiation.

I hope this video was helpful. Thanks for watching, and happy studying!

## Implicit Differentiation Practice Questions

Question #1:

Assume that $$y$$ is a function of $$x$$. Use implicit differentiation to find $$\frac{dy}{dx}$$ given the equation $$4xy+y=x^2$$.

$$\frac{dy}{dx}=\frac{2x-4y}{4x+1}$$
$$\frac{dy}{dx}=2x-4y$$
$$\frac{dy}{dx}=2x-xy$$
$$\frac{dy}{dx}=\frac{2x-4}{4x+y}$$

The first step of implicit differentiation is taking the derivative of all terms on both sides.
$$\frac{d}{dx}4xy+\frac{d}{dx}y=\frac{d}{dx}x^2$$

From here, each term can be handled individually. To find the derivative of $$4xy$$, notice that the product rule is needed. This gives us:
$$\frac{d}{dx}4xy=4x⋅\frac{d}{dx}y+y⋅\frac{d}{dx}4x$$

$$\frac{d}{dx}y$$ is equivalent to saying $$\frac{dy}{dx}$$, and $$\frac{d}{dx}4x$$ is simply $$4$$, so the derivative of $$4xy$$ is
$$\frac{d}{dx}4xy=4x\frac{dy}{dx}+4y$$

Moving along to the second term, $$\frac{d}{dx}y$$ is again equivalent to $$\frac{dy}{dx}$$.
The final term, on the right side, is easy to differentiate.
$$\frac{d}{dx}x^2=2x$$

Putting all of the terms’ derivatives together,
$$\frac{d}{dx}4xy+\frac{d}{dx}y=\frac{d}{dx}x^2$$ ⟹ $$4x\frac{dy}{dx}+4y+\frac{dy}{dx}=2x$$

The goal of this problem is finding $$\frac{dy}{dx}$$, which can now be solved for by using some algebra.
$$4x⋅\frac{dy}{dx}+4y+\frac{dy}{dx}=2x$$
$$4x⋅\frac{dy}{dx}+\frac{dy}{dx}=2x-4y$$
$$\frac{dy}{dx}(4x+1)=2x-4y$$
$$\frac{dy}{dx}=\frac{2x-4y}{4x+1}$$

Question #2:

Assume that $$y$$ is a function of $$x$$. Use implicit differentiation to find $$\frac{dy}{dx}$$ given the equation $$\text{sin}(y)++x^2y^3=\frac{\pi}{4}$$.

$$\frac{dy}{dx}=-\text{cos}(y)+6xy^2$$
$$\frac{dy}{dx}=\frac{6xy^2}{\text{cos}(y)}$$
$$\frac{dy}{dx}=\frac{\text{cos}(y)-2xy}{3x^2y}$$
$$\frac{dy}{dx}=\frac{-2xy^3}{\text{cos}(y)+3x^2y^2}$$

Step one of implicit differentiation is writing in $$\frac{d}{dx}$$ for all terms.
$$\frac{d}{dx}\text{sin}(y)+\frac{d}{dx}x^2y^3=\frac{d}{dx}(\frac{\pi}{4})$$

It is easy to see that the right side becomes 0, so we can write that down now.
$$\frac{d}{dx}\text{sin}(y)+\frac{d}{dx}x^2y^3=0$$

From here, work one term at a time through the left side. The derivative of $$\text{sin}(y)$$ can be found by changing to $$\text{cos}(y)$$ and applying the chain rule to $$y$$, since it is a function inside the sine function.
$$\frac{d}{dx}\text{sin}(y)=\text{cos}(y)⋅\frac{d}{dx}y=\text{cos}(y)\frac{dy}{dx}$$

Now, the derivative of $$x^2y^3$$ can be found using the product rule and then the chain rule.
$$\frac{d}{dx}x^2y^3=x^2⋅\frac{d}{dx}y^3+y^3⋅\frac{d}{dx}x^2$$

The derivative of $$x^2$$ is simply $$2x$$. The derivative of $$y^3$$ can be found by remembering that $$y$$ is still a function of $$x$$, so we apply the chain rule to see that
$$\frac{d}{dx}y^3=3y^2⋅\frac{d}{dx}y=3y^2\frac{dy}{dx}$$

So then
$$\frac{d}{dx}x^2y^3=x^2⋅3y^2\frac{dy}{dx}+2xy^3$$

Putting together the derivatives from each term of the original problem, we see that
$$\frac{d}{dx}\text{sin}(y)+\frac{d}{dx}x^2y^3=\frac{d}{dx}(\frac{\pi}{4})$$ ⟹ $$\text{cos}(y)\frac{dy}{dx}+3x^2y^2\frac{dy}{dx}+2xy^3=0$$

All that’s left now is getting $$\frac{dy}{dx}$$ by itself. Be careful in these final steps because the algebra can get messy.
$$\text{cos}(y)⋅\frac{dy}{dx}+3x^2y^2⋅\frac{dy}{dx}+2xy^3=0$$
$$\frac{dy}{dx}(\text{cos}(y)+3x^2y^2)=-2xy^3$$
$$\frac{dy}{dx}=\frac{-2xy^3}{\text{cos}(y)+3x^2y^2}$$

Question #3:

Assume that $$y$$ is a function of $$x$$. Use implicit differentiation to find $$\frac{dy}{dx}$$ given the equation $$\text{ln}(xy^2)=4x$$.

$$\frac{dy}{dx}=\frac{4xy-y}{2x}$$
$$\frac{dy}{dx}=4-\frac{1}{xy^2}$$
$$\frac{dy}{dx}=4xy^2-y^2$$
$$\frac{dy}{dx}=\frac{2x-2y}{xy}$$

Start by writing in $$\frac{d}{dx}$$ for all terms.
$$\frac{d}{dx}\text{ln}(xy^2)=\frac{d}{dx}4x$$

Beginning from the left side, what is the derivative of $$\text{ln}(xy^2)$$? For this term, we need to use the chain rule twice. The first use is because there is a function within the natural log, $$xy^2$$.
$$\frac{d}{dx}\text{ln}(xy^2)=\frac{1}{xy^2}⋅\frac{d}{dx}xy^2$$

The derivative of a natural log is 1 divided by the argument, multiplied by the argument’s derivative. The argument’s derivative can be found now by first employing the product rule, then, because $$y$$ is a function of $$x$$, using the chain rule a second time.
$$\frac{d}{dx}xy^2=x⋅\frac{d}{dx}y^2+y^2⋅\frac{d}{dx}x$$

Again, $$y$$ itself is a function, so in taking the derivative of $$y^2$$, it is necessary to drop the power, multiply by $$2$$, and multiply by $$y$$’s derivative.
$$\frac{d}{dx}y^2=2y\frac{dy}{dx}$$

Time to plug this back into the above expressions.
$$\frac{d}{dx}xy^2=2xy\frac{dy}{dx}+y^2$$
$$\frac{d}{dx}\text{ln}(xy^2)=\frac{1}{xy^2}⋅(2xy\frac{dy}{dx}+y^2)$$

This is the left side of the equation we were given. The right side, $$4x$$, can be differentiated on its own very easily. Putting left and right sides together after taking these derivatives, we have
$$\frac{1}{xy^2}⋅(2xy\frac{dy}{dx}+y^2)=4$$

Now, solve for $$\frac{dy}{dx}$$.
$$\frac{1}{xy^2}⋅(2xy\frac{dy}{dx}+y^2)=4$$
$$2xy\frac{dy}{dx}+y^2=4xy^2$$
$$2xy\frac{dy}{dx}=4xy^2-y^2$$
$$\frac{dy}{dx}=\frac{4xy^2-y^2}{2xy}$$
$$\frac{dy}{dx}=\frac{4xy-y}{2x}$$

Question #4:

Assume that $$y$$ is a function of $$x$$. Use implicit differentiation to find $$\frac{dy}{dx}$$ given the equation $$\frac{3y}{\text{sin}(y)}=x$$.

$$\frac{dy}{dx}=\frac{\text{cos}(y)}{3\text{cos}(y)-y\text{sin}(y)}$$
$$\frac{dy}{dx}=x\text{cos}(y)+\frac{1}{3}$$
$$\frac{dy}{dx}=\frac{\text{sin}^2(y)}{3\text{sin}(y)-3y\text{cos}(y)}$$
$$\frac{dy}{dx}=\frac{3\text{sin}(y)-3y\text{cos}(y)}{\text{sin}^2(y)}$$

First, write $$\frac{d}{dx}$$ before each term.
$$\frac{d}{dx}(\frac{3y}{\text{sin}(y)})=\frac{d}{dx}x$$

On the left side, we find the derivative by using the quotient rule.
$$\frac{d}{dx}(\frac{3y}{\text{sin}(y)})=\frac{\text{sin}(y)⋅\frac{d}{dx}3y-3y⋅\frac{d}{dx}\text{sin}(y)}{\text{sin}^2(y)}$$

The derivative of $$3y$$ is $$3\frac{dy}{dx}$$. The derivative of $$\text{sin}(y)$$ is $$\text{cos}(y)\frac{dy}{dx}$$. Therefore, the derivative simplifies to
$$\frac{d}{dx}(\frac{3y}{\text{sin}(y)})=\frac{3\text{sin}(y)\frac{dy}{dx}-3y\text{cos}(y)\frac{dy}{dx}}{\text{sin}^2(y)}$$

On the right side, the derivative of $$x$$ is simply 1. With this knowledge of both sides’ derivatives,
$$\frac{d}{dx}(\frac{3y}{\text{sin}(y)})=\frac{d}{dx}x$$ ⟹ $$\frac{3\text{sin}(y)\frac{dy}{dx}-3y\text{cos}(y)\frac{dy}{dx}}{\text{sin}^2(y)}=1$$

Now, solve for $$\frac{dy}{dx}$$ by factoring it out of the numerator, then dividing everything away.
$$\frac{\frac{dy}{dx}(\text{sin}(y)-3y\text{cos}(y))}{\text{sin}^2(y)}=1$$
$$\frac{dy}{dx}(3\text{sin}(y)-3y\text{cos}(y))=\text{sin}^2(y)$$
$$\frac{dy}{dx}=\frac{\text{sin}^2(y)}{3\text{sin}(y)-3y\text{cos}(y)}$$

Question #5:

Assume that $$y$$ is a function of $$x$$. Use implicit differentiation to find $$\frac{dy}{dx}$$ given the equation $$\sqrt{x^3y^2}=6x^2$$.

$$\frac{dy}{dx}=\frac{24\sqrt{x}-3y}{2x}$$
$$\frac{dy}{dx}=\frac{12x-\sqrt{x}}{2y}$$
$$\frac{dy}{dx}=12xy-\frac{3x^2}{2}$$
$$\frac{dy}{dx}=\frac{1}{12}x-6x^2y$$

First, write $$\frac{d}{dx}$$ in front of each term.
$$\frac{d}{dx}\sqrt{x^3y^2}=\frac{d}{dx}6x^2$$

To find the derivative of the term on the left-hand side, apply the chain rule.
$$\frac{d}{dx}\sqrt{x^3y^2}=\frac{1}{2\sqrt{x^3y^2}}⋅\frac{d}{dx}x^3y^2$$

The derivative of $$x^3y^2$$ can then be found using the product rule.
$$\frac{d}{dx}x^3y^2=x^3⋅\frac{d}{dx}y^2+y^2⋅\frac{d}{dx}x^3$$

The derivative of $$x^3$$ is of course $$3x^2$$, and we use implicit differentiation and the chain rule to find that the derivative of $$y^2$$ is $$2y\frac{dy}{dx}$$. So the derivative of $$x^3y^2$$ simplifies to
$$\frac{d}{dx}x^3y^2=2x^3y\frac{dy}{dx}+3x^2y^2$$.

Plug this back into the derivative of the first term:
$$\frac{d}{dx}\sqrt{x^3y^2}=\frac{1}{2\sqrt{x^3y^2}}(2x^3y\frac{dy}{dx}+3x^2y^2)$$

Now, find the derivative of the term on the right-hand side of the equation.
$$\frac{d}{dx}6x^2=12x$$

Putting both sides together now, we have
$$\frac{1}{2\sqrt{x^3y^2}}(2x^3y\frac{dy}{dx}+3x^2y^2)=12x$$.

All that’s left to do is solve for $$\frac{dy}{dx}$$.
$$\frac{1}{2\sqrt{x^3y^2}}(2x^3y\frac{dy}{dx}+3x^2y^2)=12x$$
$$(2x^3y\frac{dy}{dx}+3x^2y^2)=12x(2\sqrt{x^3y^2})$$
$$2x^3y\frac{dy}{dx}=24x\sqrt{x^3y^2}-3x^2y^2$$
$$\frac{dy}{dx}=\frac{24x\sqrt{x^3y^2}-3x^2y^2}{2x^3y}$$

This result can be simplified a bit further by pulling an $$x$$ and a $$y$$ from the radical, and cancelling some $$x$$’s and $$y$$’s in the numerator and denominator.
$$\frac{dy}{dx}=\frac{24x^2y\sqrt{x}-3x^2y^2}{2x^3y}$$
$$\frac{dy}{dx}=\frac{24\sqrt{x}-3y}{2x}$$

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by Mometrix Test Preparation | Last Updated: December 26, 2023