Implicit derivatives

You already know how to find a derivative of function where one variable is being expressed in terms of another, for example, $f(x) = 6x^2 + 9x+10$. There $f(x)$ is permutable with $y$, i.e you can write $y= 6x^2 + 9x+10$.

But there are many other functions that can't set up exactly like this. For example: $x^2 + y^2 = 25$ $-$ it is the equation of a circle with it's center at the origin and a radius of $5$. How can you write the derivative of that function? In order to do so, you need to know a technique called implicit differentiation, which you will study in this topic.

Implicit derivative

Let's consider an example we were just discussing: $x^2 + y^2 = 25$. In order to find $y'$ we could solve for $y$ in terms of $x$, and then take the derivative as we already know how:

$$x^2 + y^2 = 25 \ \ \Rightarrow \ \ y = \pm\sqrt{25 - x^2} \ \ \Rightarrow \ \ y' = \left(\pm\sqrt{25 - x^2} \right)'$$But there are many cases, where this can be difficult or even impossible to do. Implicit differentiation will help us to find the derivative of such function without having to solve for one variable in terms of another. We can simply differentiate with respect to one specific variable. Usually we differentiate with respect to $x$, so let's do that with this function, and then we can solve the resulting expression for $y'$ and $\dfrac{dy}{dx}$, which we are looking for.

First, we take the derivative of both sides with respect to $x$:

$$\dfrac{d}{dx} (x^2 + y^2) = \dfrac{d}{dx} (25)$$Symbol $\dfrac{d}{dx}$ means that we take the derivative with respect to $x$.

Second, we know that the derivative of a sum is the sum of derivatives, so we get

$$\dfrac{d}{dx} x^2 + \dfrac{d}{dx}y^2= \dfrac{d}{dx} (25)$$

Notice that we are considering $y$ as a function of $x$.

When we take the derivative of $x^2$ with respect to $x$, we get $2x$. Let's consider the second term: $\dfrac{d}{dx}y^2$. In this case, the situation is slightly different, because we are not taking the derivative with respect to $y$. We find the derivative with respect to $x$, and we are thinking of $y$ as a function that depends on $x$. So, for the second term we need to use the chain rule. We know that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function: $f'(g(x)) = f'(g(x))\cdot g'(x)$. In the second term we take the derivative of $y^2$ with respect to $y$, thinking of this exponent as the outer function operating on $y$, and then we also imagine $y$ as some other expression of $x$, so we multiply by the derivative of $y$ with respect to $x$. Thus, we get that $\dfrac{d}{dx}y^2 = 2y \cdot y'$.

Considering that a derivative of a constant is zero, in the end we get:

$$2x + 2y \cdot y' = 0$$We want to find $y'$, so let's solve for it:

$$2y\cdot y' = -2x \ \ \Rightarrow \ \ y' = -\dfrac{2x}{2y} = -\dfrac{x}{y}$$Thus, this is the derivative of the origin function with respect to $x$.

A second example

Let's look at a more complex example:

$$2x^3 + x^3 \cdot y + y^3 =8$$Suppose we want to find $y'$. We will take the derivative of every term with respect to $x$ again:

$$\dfrac{d}{dx} (2x^3 + x^3 \cdot y + y^3 ) =\dfrac{d}{dx}(8) \\ \ \\ \dfrac{d}{dx} (2x^3) + \dfrac{d}{dx} (x^3 \cdot y) + \dfrac{d}{dx} (y^3) = \dfrac{d}{dx} (8)$$For the first term we get $6x^2$. For the second term we need to use the chain rule, so

$$\dfrac{d}{dx} (x^3 \cdot y) = (x^3)'\cdot y + x^3 \cdot y' = 3x^2y + x^3y'$$Again, we notice, that $y$ is a function, and it's derivative is some other function, which is what we are looking for, so we write $y'$.

For the third term $\dfrac{d}{dx} (y^3)$ we need to use the chain rule: we take the derivative with respect to $y$ first, ignoring what's inside $y$, and get $3y^2$, but then we multiplying by the derivative of $y$ with respect to $x$ and get $y'$. So, in the end we get $3y^2 \cdot y'$. Thus, we get

$$6x^2 + 3x^2y + x^3y' + 3y^2y' = 0$$Hence,

$$x^3y' +3y^2 y' = - 6x^2 - 3x^2y \\ \ \\ y'(x^3 + 3y^2) = -6x^2 -3x^2y \\ \ \\ y' = -\dfrac{6x^2 +3x^2y}{x^3 + 3y^2}$$

Conclusion

In this topic, we learned about implicit derivatives. Now we know how to find a derivative of function where one variable is not being expressed in terms of another, for example, $x^3+y^3 = 8$ or $\cos (xy) = 0$.