# Implicit Differentiation – Explicitly Explained

Hey guys. Welcome to this video on implicit differentiation.

So, far you have probably been able to find derivatives of functions like:
$$y=4(3x^{2}+4x)^{2}$$ and $$R=7x^{3}\cdot 5x^{8}$$

In the case of each of these functions $$y$$ is equal to some function of $$x$$, and if we know $$x$$ we then can solve for $$y$$ directly. When this is the case we call this an explicit function.
Now an implicit function, on the other hand, is when you have a function of $$y$$ and $$x$$ that is equal to something else. Like in these functions for example: $$x + y = 1$$, $$2xy= 1$$, $$x^{2}+y^{2}=16$$
Well, these are simple enough. We can just solve for $$y$$ then take the derivative with respect to $$x$$ and get $$y’=1$$ and $$y’=-(\frac{1}{2}x^{2})$$. And this is true, we can just solve for $$y$$ then take the derivative with respect to $$x$$, with these two functions. The problem however, is that this can’t be done with every function. So, then in order take the derivative, we will need to leave it as is, and take the derivative that way. This is what we call implicit differentiation, and implicit differentiation is just an application of the chain rule. So, as you try and navigate through thinking what on earth is implicit differentiation.. Just think “chain rule.”

Let’s look at an example and I will show you what I mean. We’ve already found the derivative of this one, but let’s find the derivative using implicit differentiation.

$$x^{2}+y^{2}=16$$ $$(\frac{d}{dx})x^{2}+(\frac{d}{dy})y^{2}=(\frac{d}{dx})16$$

What is happening with our $$(\frac{d}{dy})y^{2}$$ is what is really intriguing. Because $$y$$ is a function of $$x$$, and we are taking the derivative of $$2y$$ with respect to $$x$$, we must apply the chain rule. So, really what this looks like is: $$\frac{d}{dx}[y(x)]^{2}$$ which is equal to $$2y\cdot \frac{dy}{dx}$$. So, when we apply this we get:

$$2x+2y(\frac{dy}{dx})=0$$

Now, we have this derivative of $$y$$ with respect to $$x$$ in our equation. So, all we have to do at this point is solve for our $$\frac{dy}{dx}$$.

Let’s subtract $$2x$$ from both sides, then divide by $$2y$$ to give us:

$$\frac{dy}{dx}=-\frac{x}{y}$$

Now, this is different than what you are probably use to seeing when you have $$y$$ explicitly defined as a function of $$x$$. But, now what we have is is our derivative in terms of $$x$$ and $$y$$, instead of just $$x$$. Like, in our explicitly defined function.

Alright, let’s work another problem using implicit differentiation.

Assume $$y$$ is a function of $$x$$. Find $$\frac{dy}{dx}$$ for $$x=7\sqrt{x^{2}+y^{2}}$$.

So, to start, we need to take the derivative with respect to $$x$$ of both sides.

$$\frac{d}{dx}(x)=\frac{d}{dx}(7+\sqrt{x^{2}+y^{2}})$$

Now, our left side is pretty simple, but to make our right side a bit simpler I am going to rewrite the square root part.

$$=\frac{d}{dx}(x)=\frac{d}{dx}(7+(x^{2}+y^{2})^{\frac{1}{2}})$$

Now, we just take the derivative. To the right of our equal sign we can take the derivative of 7 and our $$(x^{2}+y^{2})^{\frac{1}{2}}$$ separately, because it is addition. But with our $$(x^{2}+y^{2})^{\frac{1}{2}}$$, we will need to use the chain rule, then technically apply the chain rule again when we go to take the derivative of $$y$$ in our inside function with respect to $$x$$.

$$1=\frac{1}{2}(x^{2}+y^{2})^{-\frac{1}{2}}\cdot 2x+2y(\frac{dy}{dx})$$

So, at this point we’ve taken the derivative, and now we just need to do some algebra to get $$\frac{dy}{dx}$$ by itself. I’m going to go ahead and rewrite this by placing our $$(x^{2}+y^{2})^{-\frac{1}{2}}$$in the denominator, and factoring out a 2 from our $$2x+2y(\frac{dy}{dx})$$.

$$1=\frac{(\frac{1}{2}\cdot 2(x+y\cdot \frac{dy}{dx}))}{\sqrt{x^{2}+y^{2}}}$$

Our $$\frac{1}{2}$$ and 2 will cancel out, and we can multiply both sides by $$\sqrt{x^{2}+y^{2}}$$.

$$\sqrt{x^{2}+y^{2}}=x+y\cdot \frac{dy}{dx}$$

Now, we just need to subtract an $$x$$ from both sides.

$$\sqrt{x^{2}+y^{2}}-x=y\cdot \frac{dy}{dx}$$

Lastly, we need to divide by sides by $$y$$.

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

I hope this video better helped you to understand implicit differentiation. For further help, be sure to check out more of our videos by subscribing to our channel below.

## 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.
$$\frac{dy}{dx}=\frac{24x^2y\sqrt{x}-3x^2y^2}{2x^3y}$$
$$\frac{dy}{dx}=\frac{24\sqrt{x}-3y}{2x}$$