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Implicit Differentiation

This is what put an end to my mathematics career in college. I thought "implicit differentiation" meant the differentiation was implicit. I could not figure out what that meant. In fact, there is nothing implicit about the differentiation. It works the chain rule to death, but it is very straight forward and simple.

Implicit differentiation means differentiating functions that are expressed implicitly. The functions are implicit, not the differentiation.

Explicit and Implicit Functions

An explicit function is the answer to an algebra problem that starts "Solve for y" (or whatever variable). That answer has whatever variable you solved for naked on one side (usually the left) of the equals sign and everything else on the other side of the equals sign. By naked, I mean it does not have a constant to be multiplied by, it is not expressed as a power or a root (which amounts to the same thing), it is not added or subtracted from anything, and it is not wrapped in a function like sin or tan.

An implicit function looks like the algebra problem whose solution is an explicit function.

Often, there is only a little algebra between an explicit function and an implicit function:

{$$ \eqalign { \frac{1}{2} y &= x + 2 \text{ implicit} \cr y &= 2x + 4 \text{ explicit}} $$}

A fairly famous implicit function is:

{$$ x^2 + y^2 = 1 $$}

which of course really is the implicit form of two functions of y (or x if you want to go that way) because solving for y does not provide a unique function:

{$$ y = \pm \sqrt{x^2 + 1} $$}

But students who point this out in mid-lecture are universally regarded as a pain in the butt.

Why differentiate implicit functions

In some cases the derivative can be found either way. But one form or another may suggest the involvement of hairy algebra. For example, the explicit forms of the circle functions, with addition in a root, suggests many lines of algebra.

In other cases, one form or the other must be chosen because it is not known how to take the derivative or it is not known how to solve the equation.

Demonstration

To implicitly differentiate for y as a function of x:

{$$ x^2y = 1 $$}

Of course we could solve for y, but that is the next to last step in implicit differentiation. Instead we differentiate through:

{$$ \begin{align} {d \over dx} \left( x^2y \right) &= {d \over dx} 1 \cr ({d \over dx}x^2)y + x^2({d \over dx}y) &= 0 \tag{product rule} \cr 2xy + x^2{dy \over dx} &= 0 \cr {dy \over dx} &= \frac{-(2xy)}{x^2} \cr \cr \therefore {dy \over dx} &= \frac{-2y}{x}\end{align} $$}

If you know that the powers rule is valid for negative integer powers (at least) it is absurdly simple to solve the equation explicitly in the first place (which can be done as a check), but the purpose here is to illustrate implicit differentiation. There is nothing profound or acane going on here. The calculus part of the problem here is to see the multiplication rule is called for (and to remember it).

A Classic Demonstration

I would like to bring something original to the table here, but this example of implicit differentiation occurs in every textbook and in the lecture UTubes I have seen.

I am going to differential the equation of a circle. I am not going to fool with the plus or minus of the roots, because I am going leave it in the implicit form which will answer for both the top and the bottom of the circle.

{$$ \begin{align} x^2 + y^2 &= 1 \tag{circle} \cr {d \over dx}(x^2 + y^2) &= {d \over dx}1 \cr {d \over dx}(x^2) + {d \over dx}(y^2) &= 0 \cr 2x + 2y{dy \over dx} &=0 \tag{chain rule} \cr {dy \over dx} &= \frac{-2x}{2y} \cr \therefore {dy \over dx} &= - \frac{x}{y} \end{align} $$}

About the only thing esoteric here is I invoked the chain rule. Although using the chain rule with very complex expressions may require acres of board or reams of paper with many intermediate variables, simple cases like this should be written out directly, to the point that it becomes automatic.

Now about the result: this result really does tell us the slope of the tangent to the circle at any specified point on the circle. Simply substitute the particular x value and y value, and you are done. The signs are gauranteed to be right from the fact that the point is on the circle. The slope will be negative in the first and third quadrants where x and y have the same sign, and will be positive in the second and fourth quadrants where x and y have different signs.

Many times it will be desirable to express the derivative entirely in terms of x (or whatever the independent variable was). This can be done by returning to the implicit function, solving it explicitly, and substituting.

{$$ \begin{align} {dy \over dx} &= - \frac{x}{y} \tag{our result} \cr x^2 + y^2 = 1 & \tag{implicit function} \cr y=\pm \sqrt{1 - x^2} & \tag{explict functions} \cr \therefore {dy \over dx} &= - \frac{x}{\sqrt{1 - x^2}} \tag{substituting + root} \cr \therefore {dy \over dx} &= \frac{x}{\sqrt{1 - x^2}} \tag{substituting - root} \end{align} $$}

Now the sign is not automatically adjusted. It is necessary to know which half of the circle you are working with. The practical needs of a practical problem will determine where to leave off the solution.

Implict Differentiation of Inverse Functions

A major use of implicit differentiation is to differentiate inverse functions when the derivative of the original function is known.

As an example, I will demonstrate finding the derivative of arcsec(x).

Derivative of arcsec(x)

O θ D B tan θ 1 sec θ A C


A Pythagorean Identity {$$ \begin{align} tan^2(x) + 1 &= sec^2(x) \cr tan^2(x) &= sec^2(x) -1 \cr tan(x) &= \pm \sqrt{sec^2(x) -1} \end{align} $$}

Given: {$f'(sec(x)) = sec(x)tan(x)$} find {$ f'(arcsec(x))$}.

Arcsec is the inverse function of sec sometimes written {$ \sec^{-1} $}. The derivative of sec is given just in case you have not yet derived it.

The essence of inverse of a function is that is undoes whatever the function does so

{$$ g = f^{-1} \iff g \left( f(x) \right) = x $$}

An important point of notation here is that the -1 does not mean recipricol.

{$$ f^{-1}(x) \ne \frac{1}{f(x)} $$}

This is not consistent with other uses of exponent numbers on functions, but it is very common.

Now the problem with trig functions is that they are periodic. That is why we can use the unit circle: because if we twist the angle far enough, we come back to the same place, and we can keep on twisting as long as we please. When we work only from the values of trig functions, we do not have access to the information about how much we twisted before we stopped at a particular point on the unit circle. So we cannot have a function that really undoes secant. To have an inverse, we have to promise to twist only so far as we can without repeating a value of secant. In the case of secant we have to promise not to twist quite so much as π and not to twist backwards at all.

We can have an inverse of this function:

{$$ g(x) = sec(x), x \in [0, \pi), x \ne {\pi \over 2} $$}

That is we require x to be in the first or second quadrant.

Now we can consider the inverse:

{$$ \begin{align} sec(arcsec(x)) &= x \tag{definition of inverse} \cr \text{Let } y = arccos(x)& \cr sec(y) &= x \cr {d \over {dx}} \left(sec(y)\right) &= {d \over {dx}} (x) \end{align} $$}

I am taking the derivative with respect to x of both sides. Remembering that y is a function of x, the chain rule is invoked on the left side.

{$$ \begin{align} {d \over {dx}} \left(sec(y)\right) &= {d \over {dx}} (x) \cr {d \over {dx}}\left(sec(y)\right){d \over {dx}}y &= 1 \tag{chain rule} \cr {d \over {dy}}\left(sec(y)\right)= sec(y)tan(y) & \tag{given} \cr sec(y)tan(y){d \over {dx}}y &= 1 \cr {d \over {dx}}y &= \frac{1}{sec(y)tan(y)} \cr {d \over {dx}}y &= \frac{1}{xtan(y)}\end{align} $$}

Now the claim is that this quantity must be positive. I do not doubt it, but I have yet to see a rigorous defense of it. Often at this point there is waving at the graph, which upon inspection does support the claim. An algebraic approach is offered in the lemma.

{$$ \begin{align}{d \over {dx}}y &= \frac{1}{xtan(y)}\end{align} $$}

-3 -2 -1 1 π/2 2 π -3 -2 -1 0 1 π/2 2 π 4 5 sec x arcsec x

{$$ {d \over {dx}}y = \frac{1}{|x|\sqrt{sec^2y-1}} $$}

Here we appealed to the Pythagorean identity {$ tan^2y + 1 = sec^2y $}, which yields {$ tan^2y = sec^2y - 1 $}, taking the root of both sides, but choosing only the positive root, which paired with the absolute value of x will ensure that the value is non-negative. Then, {$ sec(y) = x $}, which is where we began, is substituted in.

{$$ \therefore \boxed{{d \over {dx}}y = \frac{1}{|x|\sqrt{x^2-1}} } $$}

Lemma

To show: {$ sec(y)tan(y) \ge 0 $} when {$ sec(y) = x $} and {$ 0 \le y \le \pi $}, {$ y \ne {\pi \over 2} $}.

{$$ \begin{align} sec(y)tan(y) &= sec(y)tan(y) \cr &= \left( \frac{1}{cos (y)} \right) tan(y) \cr &= \left( \frac{1}{cos (y)} \right) \frac{sin(y)}{cos (y)} \cr &= \frac{sin(y)}{cos^2 (y)} \end{align} $$}

Now, y can only assume values in the first and second quadrant where sin is always non-negative:

{$$ \begin{align} sin(y) & \ge 0 \tag{domain of y} \cr cos^2(y) & \ge 0 \cr \frac{sin(y)}{cos^2(y)} & \ge \frac{0}{cos^2(y)} \cr \frac{sin(y)}{cos^2(y)} & \ge 0 \end{align} $$}

So: {$$ \begin{align} sec(y)tan(y) &= \frac{sin(y)}{cos^2 (y)} \ge 0 \cr \therefore sec(y)tan(y)&\ge 0 \end{align} $$}

In short, it comes down to sin(y) and sin is always non-negative in the first and second quadrant.

Note: some texts define the domain of y differently in which case this argument is not valid.


Sources:

  1. FooPlot | Online graphing calculator and function plotter
  2. FooPlot | Online graphing calculator and function plotter

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August 06, 2017

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