Introduction to Implicit Derivatives:
We have been applying differentiation to functions where one variable is being expressed in terms of another. In these functions, f(x) is interchangeable with “y” and some “x” terms will follow.
But there are plenty of other functions that aren't set up exactly like this for example:
$$ x^2 + y^2 = 9 $$
Now, the equation is a function which containes more than one variable, such type of a function is called as implicit function.
It is the equation of a circle with its center at the origin and a radius of 3.
How would we take the derivative of this ?
In order to do it we need to use a technique called Implicit differentiation.
Implicit Differentiation
Consider the same example of implicit derivative. We were just discussing we know that we could solve for “y” in terms of “x”.
The implicit function for differentiation is given as:
$$ y= \sqrt {9-x^2} $$
Then take the derivative as we already know how but this can be difficult and time-consuming and sometimes it's not even possible
$$ (\sqrt {9-x^2})' $$
Implicit differentiation will allow us to take the derivative of the function without having to solve for one variable in terms of another instead.
We can just leave everything as it is and simply differentiate with respect to one specific variable.
We typically differentiate with respect to “x”. In general we will solve for y
So let's do that to this function and then we can solve for y' or differentiate with respect to x
First thing is to take the derivative of both sides with respect to “x”
$$ \frac{d}{dx}(x^2+y^2) = \frac{d}{dx}(9) $$
Or,
$$ \frac{d}{dx}x^2 + \frac{d}{dx}y^2 = \frac{d}{dx}(9) $$
We know the derivative of any constant is zero, So the Right hand side will become zero.
$$ \frac{d}{dx}x^2 + \frac{d}{dx}y^2 = 0 $$
Now, derivative of x2 w.r.t “x” can be calculated. But the derivative of y2 w.r.t “x” is difficult because we can’t take the derivative of one variable which is a function in terms of another.
Thus we have to use Chain-Rule there. Remember the concept of Chain-Rule:
$$ [f(g(x))]' = f'(g(x)) \cdot g'(x) $$
We have a composite function y2, then according to chain-rule we can take the derivative of outer function w.r.t to itself times to derivative of inner function respectively.
So, we will get
$$ 2x + \frac{d}{dy}y^2 \cdot \frac{d}{dx}(y) = 0 $$
Now, taking the derivative of y2 w.r.t “y” is easy
$$ 2x + 2y\frac{dy}{dx} = 0 $$
Or, we can also write
$$ 2x + 2y \cdot y' = 0 $$
And in the last step we have to solve for y'
So,
$$ 2y \cdot y' = -2x $$ $$ y' = -\frac{2x}{2y} $$ $$ y' = -\frac{x}{y} $$
So, this is the implicit derivative of our implicit function with respect to x.
So that's all there is about implicit differentiation, it seems tricky at first but once you get the algorithm down, it becomes easy to differentiate with respect to x.
As we are just applying things that we already know like the chain rule the only trick is to remember that if solving for y’ it means that “y” is some function in terms of “x” and anytime we take the derivative of “y”, if there are other things operating on “y”.
We will have to use the chain rule which will produce the y' terms then we end up solving.