The Differentials of the Inverse Tangent and Inverse Cotangent Functions

Section 6.6 The Differentials of the Inverse Tangent and Inverse Cotangent Functions

Since we want to find a function whose derivative is the Witch of Agnesi and we are pretty sure that each of the branches of the arctangent multifunction takes the shape of the Witch’s antiderivative it is tempting to assume that the branch of the arctangent shown below is the function we are looking for.

In fact, this is correct, but we must be careful. Reasoning from pictures, as we’ve just done, can be very useful. but the most we can hope for when reasoning from pictures is an intuitive feel for the problem. We always need to confirm our intuition analytically. Always.

So we will verify, analytically, our conjecture that the derivative of the inverse tangent function is the Witch of Agnesi. From our definition of \(\inverse\tan(x)\) we know that \(y=\inverse\tan(x)\) precisely when \(\tan(y)=x\text{.}\) Differentiating \(\tan(y)=x\) gives:

\begin{align*} \dx(\tan(y)) \amp = \dx{x}\\ \sec^2(y)\dx{y} \amp = \dx{x}\\ \dx{y}\amp = \frac{1}{\sec^2(y)}\dx{x}. \end{align*}

This is a correct formula for the differential of the arctangent function. However, there are two problems: (1) it does not seem to match our conjecture, and (2) even if it is correct this formula will not be very useful to us in this form since we have \(\dx{y}\) in terms of \(y\) itself. We can address both of these problems by finding \(\dx{y}\) in terms of \(x\) and \(\dx{x}\text{.}\)

Referring to the diagram above, recall from Section 6.1 that \(\tan(y)=\frac{x}{1}=x\) so \(y=\inverse\tan(x)\text{.}\) The hypotenuse in the diagram above is \(\sec(y)\) so from the Pythagorean Theorem we see that \(\sec^2(y)=1+x^2\text{.}\) Thus

\begin{equation*} \dx{y}=\frac{1}{\sec^2(y)}\dx{x}=\frac{1}{1+x^2}\dx{x}, \end{equation*}

\begin{equation*} \dfdx{y}{x}=\frac{1}{1+x^2} \end{equation*}

So our intuition was correct. The Witch is indeed the derivative of the arctangent function. A similar derivation can be used for the arccotangent.

Problem 6.6.1.

(a)

The inverse cotangent function is defined as \(y=\inverse\cot(x)\) if and only if \(\cot(y)=x\text{.}\) Proceed as we did above to show that

\begin{equation*} \dx{\left(\inverse\cot(x)\right)} = \frac{-1}{1+x^2}\dx{x}. \end{equation*}

(b)

Derive the same result from the identity \(\inverse\cot(x)=\pi/2-\inverse\tan(x)\text{.}\)

Problem 6.6.2.

(a)

Show that if \(y=\inverse\tan(x)\) then \(y\) satisfies the differential equation

\begin{equation} (1+x^2)\dfdxn{y}{x}{2}+2x\dfdx{y}{x}=0.\tag{6.14} \end{equation}

(b)

The function \(y=\inverse\cot(x)\) also satisfies equation (6.14). Show this in two different ways.

By direct computation, just as you did part 6.6.2.a.
By direct computation, after first observing that \(\inverse\cot(x)=\frac\pi2-\inverse\tan(x)\text{.}\)

Problem 6.6.3.

Show that each of the following statements is true.

(a)

\(\dx{\left(\inverse\tan\left(\dfrac{1+x}{1-x}\right)\right)} = \dfrac{1}{1+x^2}\dx{x}\)

(b)

\(\dx{\left(\inverse\tan\left(\dfrac{2+x}{1-2x}\right)\right)} = \dfrac{1}{1+x^2}\dx{x}\)

(c)

\(\dx{\left(\inverse\tan\left(\dfrac{10+x}{1-10x}\right)\right)} = \dfrac{1}{1+x^2}\dx{x}\)

(d)

\(\dx{\left(\inverse\tan\left(\dfrac{2543+x}{1-2543x}\right)\right)} = \dfrac{1}{1+x^2}\dx{x}\)

(e)

This is weird isn’t it? None of the functions on the left is \(\inverse\tan(x)\) but they all have the same derivative as \(\inverse\tan(x)\text{.}\) Can you explain this?

Hint.

Let \(y_1=\inverse\tan(x)\) and \(y_2=\inverse\tan(\kappa)\text{.}\) What is \(\tan(y_1+y_2)\text{?}\)

(f)

Now show that \(\dx{\left(\inverse\tan\left(\dfrac{\kappa+x}{1-\kappa x}\right)\right)} = \dfrac{1}{1+x^2}\dx{x}\) where \(\kappa\) is an arbitrary constant.

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