Implicit Derivatives

It is possible to take derivatives of functions where the function that produces an output is not explicitly defined.

Consider the equation for a circle

\[ x^2 + y^2 = 1 \]

This isn’t really a function where we can state \( y \) in terms of \( x \) because it produces an output where each \( x \) input has two \( y \) outputs. This is because any vertical line for a valid input cuts the circle in two places.

However, we can re-arrange the function to state \( y \) more explicitly.

\[ x^2 + y^2 = 1 \]

\[ y^2 = 1 - x^2 \]

\[ y = \pm\sqrt{1 - x^2} \]

We can’t take a derivative of this because its still not a function because it has two branches - one positive and one negative - however, if we drop the \( \pm \) and pick and branch we create a semi-circular segment of the original equation which we can differentiate.

\[ y = \sqrt{1 - x^2} \]

We’ve now re-arranged the equation to state \( y \) in terms of \( x \).

The explicit approach

First let’s try and take the explicit derivative of this function (which you should already know how to do) so that we can appreciate the elegance of the implicit solution afterwards.

First re-arrange the function use a rational exponent:

\[ y = (1 - x^2)^{\frac{1}{2}} \]

Then apply the chain rule to take the derivative of the function.

\[ y’ = \frac{1}{2}(1 - x^2)^{-\frac{1}{2}}\cdot-2x \]

\[ y’ = -x(1 - x^2)^{-\frac{1}{2}} \]

Turn the now negative rational exponent into a fractional function with a square root

\[ y’ = \frac{-x}{\sqrt{1 - x^2}} \]

When static \( y \) explicitly above we found that \( y = \sqrt{1 - x^2} \) which is the demominator of this derivative. So it can be re-written as

\[ y’ = \frac{-x}{y} \]

This explicitly solves the derivative.

The implicit approach

The above approach required a lot of re-arranging and deciding on branches of original equation to take the derivative of to come to final answer. Instead, we can take the derivative of the original equation directly and implicitly take the derivative of \( y \) in terms of \( x \) in the process.

First, take the derivative of sides of the equation.

\[ x^2 + y^2 = 1 \]

\[ \frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}1 \]

Apply the sum rule to take the derivative of each of the terms in the left side of the equation

\[ \frac{d}{dx}x^2 + \frac{d}{dx}y^2 = \frac{d}{dx}1 \]

In this case the derivative of \( x^2 \) and \( 1 \) are easy. They equal \( 2x \) and \( 0 \) respectively. But how do we deal with the second variable \( y \) in this situation? We treat the extra variable as a function expressed in terms of \( x \) which we wrote out above. Every time you see \( y \) in the equation it actually implicitly stands for its definition in terms of \( x \) as \( y = \sqrt{1 - x^2} \). If that’s the case then we can just treat it as a nested function which means \( y^2 \) just means raising a function called \( y \) to power of \( 2 \) and so we can simply apply the chain rule here with the caveat that we don’t yet know what the derivative of \( y \) is so we just state it as \( y’ \) or \( \frac{d}{dx}y \) to stand in for the unknown derivative.

\[ 2x + 2y\cdot\frac{d}{dx}y = 0 \]

Now we want to re-arrange this derivative to be in terms of \( \frac{d}{dx}y \) to solve for that and give us the actual derivative that we wanted all along.

\[ 2y\cdot\frac{d}{dx}y = -2x \]

\[ y\cdot\frac{d}{dx}y = -x \]

\[ \frac{d}{dx}y = \frac{-x}{y} \]

This solves the derivative of \( y \) implicitly with much less headache the original explciit approach where we had to re-arrange the original equation and substitute that equation back into the function at the end.

Implicit derivative without substitution

You can take an implicit derivative of a function that doesn’t require you to subtitute the function for the value of \( y \) in the final derivative but instead use the value of \( y \) at a point directly. This is confusing as previous work with derivatives explored functions like \( \frac{d}{dx}x^2 = 2x \) where a single value for \( x \) was plugged into the derivative to find the rate of change at the point, for example at \( 3 \) the slope is \( 6 \). However, the following example has a derivative in terms of \( x \) and \( y \) so a point in those terms has to substituted to get the value of the slope.

This example uses the function

\[ y^3 + x^3 = 3xy \]

It’s possible to solve this in terms of \( y \) but difficult and messy, so let’s just go straight to solving implicitly by taking the deriative of this entire function. Note, that we have to idea what value of \( y \) is for this function in terms of \( x \).

\[ \frac{d}{dx}( y^3 + x^3 = 3xy ) \]

\[ 3y^2 \frac{d}{dx}y + 3x^2 = 3y + 3x\frac{d}{dx}y \]

We use the product rule \( u’v + v’u \) on the right side of the function. The first \( x \) dissapears because \( (x^1)’ = 1 \cdot x^0 = 1 \) and the quotient always remains when taking the derivative of a non constant term.

Next group the \( \frac{d}{dx}y \) terms and \( x \) / \( y \) terms.

\[ 3y^2 \frac{d}{dx}y - 3x\frac{d}{dx}y = 3y - 3x^2 \]

Factorise out of the \( \frac{d}{dx}y \).

\[ \frac{d}{dx}y(3y^2 - 3x) = 3y - 3x^2 \]

Divide by \( 3y^2 - 3x \) to separate the \( \frac{d}{dx}y \)

\[ \frac{d}{dx}y = \frac{3y - 3x^2}{3y^2 - 3x} \]

The derivative of \( y \) is therefore

\[ \frac{d}{dx}y = \frac{y - x^2}{y^2 - x} \]

Which gives a function in terms of \( x \) and \( y \) into which we can substitute values to find the derivative at that point. For example, at point \( (\frac{4}{3}, \frac{2}{3}) \) the derivative is

\[ \frac{ \frac{2}{3} - \frac{16}{9} }{ \frac{4}{9} - \frac{4}{3} } \]

\[ = \frac{6 - 16}{4 - 12} = \frac{5}{4} \]

We can therefore see that we can solve derivative problems even without ever knowing the value of a function input or output in explicit terms.