# 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(3x2 +4x)^2         and          R= 7x^3 * 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, x2 + y2 = 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’=- (1/2×2). 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.

x2 + y2 = 16

(d/dx)x^2 + (d/dx)y^2 = (d/dx)^16

What is happening with our (d/dx)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: d/dx [(y(x))]^2 which is equal to 2y*dy/dx. So, when we apply this we get:

2x + 2y (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 dy/dx.

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

dy/dx= – 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 dy/dx for x= 7 + sqrt(x2+y2).

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

d/dx(x)=d/dx(7 + sqrt(x2+y2))

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.

= d/dx(x)=d/dx(7 + (x2+y2)^½

Now, we just take the derivative. To the right of our equal sign we can take the derivative of 7 and our (x2+y2)½ separately, because it is addition. But with our (x2+y2)½, 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= ½ (x2+y2)^-½ * 2x + 2y(dy/dx)

So, at this point we’ve taken the derivative, and now we just need to do some algebra to get dy/dx by itself. I’m going to go ahead and rewrite this by placing our (x2+y2)^-½ in the denominator, and factoring out a 2 from our 2x + 2y(dy/dx).

1=(½ 2(x+y*dy/dx))/sqrt(x^2+y^2)

Our ½ and 2 will cancel out, and we can multiply both sides by sqrt(x^2+y^2).

sqrt(x^2+y^2)=x+y*dy/dx

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

sqrt(x^2+y^2) – x =y*dy/dx

Lastly, we need to divide by sides by y.

(sqrt(x^2+y^2) – x)/y =dy/dx