Recognize equivalent derivative forms

Two answers for one slope

Two students find the slope of \(y = \tan x\). One writes \(\sec^2 x\), and the other writes \(1 + \tan^2 x\), which looks different.

The chalk curve is \(\sec^2 x\). Draw the second answer, and the orange dashed curve of \(1 + \tan^2 x\) lies on the chalk curve.

Go to \(x = \frac{\pi}{3}\). Here \(\sec^2 x = 4\) and \(\tan x = \sqrt3\), so \(1 + \tan^2 x = 1 + 3 = 4\).

Drag the point. The curves lie on each other at every x, so the two answers are equivalent forms of one function.

Equivalent forms

A correct derivative can be written in many forms. Two forms are equivalent when they have the same value at every input where both have a value, so their curves lie on each other.

Drag \(x\), and the chalk curve and the dashed curve stay at the same height. Show another pair: the chalk curve becomes \(\tan^2 x\), the dashed curve becomes \(\sec^2 x - 1\), and they lie on each other again.

An identity is an equation that is true wherever both sides have a value, so it changes an expression's form and keeps its value.

Notation

1. Start with the identity for sine and cosine.

   \[\sin^2 x + \cos^2 x = 1\]

   It holds for every \(x\).

2. Divide both sides by \(\cos^2 x\), for \(\cos x \ne 0\). This gives the identity that the two curves in the figure show.

   \[\tan^2 x + 1 = \sec^2 x\]

3. Divide both sides by \(\sin^2 x\) instead, for \(\sin x \ne 0\). The chalk curve becomes \(\csc^2 x\), and the dashed curve of \(1 + \cot^2 x\) lies on it.

   \[1 + \cot^2 x = \csc^2 x\]

4. To compare two answers, write \(\tan\), \(\sec\), \(\cot\) and \(\csc\) with sine and cosine.

   \[\begin{aligned} \tan x &= \frac{\sin x}{\cos x}, & \sec x &= \frac{1}{\cos x} \\ \cot x &= \frac{\cos x}{\sin x}, & \csc x &= \frac{1}{\sin x} \end{aligned}\]

5. Two answers are equivalent if these identities and algebra turn one into the other. They are then equal wherever both have a value.

When it fails

At \(x = \frac{\pi}{4}\), \(\sec^2 x = 2\), so the curve and the line \(y = 2\) have the same value at this one input.

In the figure, the curve and the line meet only at \(x = \pm\frac{\pi}{4}\), so 2 is not equivalent to \(\sec^2 x\). One matching value does not prove that two forms are equal.

Example 1: sec² minus 1

Find \(y'\) for \(y = \tan x - x\). Write the answer in a second form.

1. The derivative of \(\tan x\) is \(\sec^2 x\), and the derivative of \(x\) is 1.

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

2. The identity \(\sec^2 x = 1 + \tan^2 x\) gives a second form.

   \[\begin{aligned} \sec^2 x - 1 &= (1 + \tan^2 x) - 1 \\ &= \cy{\tan^2 x} \end{aligned}\]

3. As a test, both forms give 1 at \(x = \frac{\pi}{4}\). The identity proves that they are equal at every \(x\), so the curves lie on each other.

4. The second form shows more. Because \(\tan^2 x\) is never negative, \(y = \tan x - x\) never falls. For \(-\frac{\pi}{2} < x < \frac{\pi}{2}\), its slope is 0 only at \(x = 0\).

Example 2: three forms of one slope

Show that the derivative of \(y = \csc x - \cot x\) equals \(\dfrac{1}{1 + \cos x}\), for \(0 < x < \pi\).

1. Differentiate with the rules. The derivative of \(\csc x\) is \(-\csc x\cot x\), and the derivative of \(-\cot x\) is \(+\csc^2 x\).

   \[y' = -\csc x\cot x + \csc^2 x\]

2. Write each term with sine and cosine, and add over the common denominator.

   \[\begin{aligned} y' &= -\frac{\cos x}{\sin^2 x} + \frac{1}{\sin^2 x} \\ &= \frac{1 - \cos x}{\sin^2 x} \end{aligned}\]

3. A tempting wrong move: Replace \(\sin^2 x\) by \(1 - \cos x\).

   \[\cr{y' = \frac{1 - \cos x}{1 - \cos x} = 1}\]

   At \(x = \frac{\pi}{3}\), \(\sin^2 x = \frac34\) but \(1 - \cos x = \frac12\), so the two are not equal. The red line at height 1 lies above the curve, which is \(\frac23\) high there.

4. Use the correct identity \(\sin^2 x = 1 - \cos^2 x\). It factors as a difference of squares.

   \[\sin^2 x = (1 - \cos x)(1 + \cos x)\]

5. Cancel \(1 - \cos x\), which is not 0 for \(x\) between 0 and \(\pi\). The orange dashed curve of this form lies on the chalk curve.

   \[\begin{aligned} y' &= \frac{1 - \cos x}{(1 - \cos x)(1 + \cos x)} \\ &= \cy{\frac{1}{1 + \cos x}} \end{aligned}\]

Practice

Two forms of the derivative of \(y = \sec x + \tan x\) are \(\sec x(\sec x + \tan x)\) and \(\dfrac{1 + \sin x}{\cos^2 x}\). Show that they are equivalent.

Write \(\sec x\) and \(\tan x\) with sine and cosine, over the common denominator \(\cos^2 x\). If the forms are equivalent, the dashed curve lies on the chalk curve.

Questions

The chalk curve is \(\sec^2 x\). The orange dashed curve is a second expression, and the orange line marks the input \(x\). Questions 1 and 2 use the figure. Question 3 needs a calculation.

1. The dashed curve is \(\tan^2 x\). The gap between the curves is the same at every \(x\), and the dashed curve lies below. Which identity does the figure show?

   • \(\sec^2 x = \tan^2 x + 1\)
   • \(\sec^2 x = \tan^2 x\)
   • \(\sec^2 x = \tan^2 x - 1\)
   • \(\sec x = \tan x + 1\)
   Answer

   \(\sec^2 x = \tan^2 x + 1\)

2. The line \(y = 2\) meets the curve \(y = \sec^2 x\) at \(x = \frac{\pi}{4}\). Does this show that \(\sec^2 x = 2\) for every \(x\)?

   • No. In the figure, the curve and the line meet only at \(x = \pm\frac{\pi}{4}\)
   • Yes. Both are 2 at \(x = \frac{\pi}{4}\)
   • Yes, because the line touches the curve
   Answer

   No. In the figure, the curve and the line meet only at \(x = \pm\frac{\pi}{4}\)

3. Which expression equals the derivative of \(y = \cot x + x\) at every \(x\) where \(\sin x \ne 0\)?

   • \(-\cot^2 x\)
   • \(\cot^2 x\)
   • \(-\tan^2 x\)
   • \(-\csc^2 x - 1\)
   Answer

   \(-\cot^2 x\)

Summary

You rewrite an answer with an identity until it matches another form, and you compare the two forms as functions.

The identities hold at every input where the functions have values, so two correct forms of a derivative have the same curve.

You use this when an answer looks different from the key, or when a simpler form shows the sign of the slope.

This is one of four skills in topic 2.10 of Which & Why: AP Calculus.