Introduction / Summary of Differentiation

[Hootenanny]

Inverse Trigonometric Derivatives (Part I of II)
Contributed by Gib Z. Edited by Hootenanny.

As already established in this thread:

\frac{d}{dx} \sin x = \cos x

\frac{d}{dx} \cos x= -\sin x

\frac{d}{dx} \tan x = \sec^2 x

Using the results of the previous posts regarding the derivatives of inverse functions, we can find the derivatives of the inverse trigonometric functions fairly trivially. However, since the trigonometric functions are not one-to-one functions on the current domain of the real numbers, we must restrict the domain to a small interval, so that their inverse relations are well defined functions. There are many possible choices we could make, and would be suitable, but we choose specific ones for convenience. We call the inverse functions of \sin x, x \in \left[-\pi , \pi\right]; \cos x, x \in \left[0,\pi\right] and \tan x, x \in \left[-\frac{\pi}{2} , \frac{\pi}{2}\right]; \arcsin x;\arccos x and \arctan x respectively.

A point to note here is that since the trigonometric functions have a restricted domain (and a restricted range), the inverse trigonometric functions will have a restricted range. A nice feature of of inverse functions it that the domain of a given function, becomes the range of the inverse function and vice-versa. For example, if we restrict \cos x to a domain of x \in \left[0,\pi\right], then the [restricted] range is, \cos x \in \left[-1,1\right]. Hence, the corresponding inverse function \arccos x has domain x \in \left[-1,1\right] and range \arccos x \in \left[0,\pi\right].

 

[Hootenanny]

Inverse Trigonometric Derivatives (Part II of II)
Contributed by Gib Z. Edited by Hootenanny.

Notation
A small note here on notation, there are two common notations for denoting inverse trigonometric functions.

\arcsin x = \sin^{-1} x

\arccos x = \cos^{-1} x

\arctan x = \tan^{-1} x

It should also be noted that the exponent does not denote the reciprocal, explicitly,

\sin^{-1} x \neq \frac{1}{\sin x} = \mathrm{cosec}\; x

\cos^{-1} x \neq \frac{1}{\cos x} = \sec x

\tan^{-1} x \neq \frac{1}{\sin x} = \cot x
​

If we now evaluate the derivatives of these [inverse] functions:

y := \arcsin x \Leftrightarrow x = \sin y \Rightarrow \frac{dx}{dy} = \cos y

\frac{dy}{dx} = \frac{1}{ \frac{dx}{dy}} = \frac{1}{\cos y} = \frac{1}{\sqrt{1-\sin^2 y}} = \frac{1}{\sqrt{1-x^2}}.

y := \arccos x \Leftrightarrow x = \cos y \Rightarrow \frac{dx}{dy} = -\sin y

\frac{dy}{dx} = \frac{1}{ \frac{dx}{dy}} = \frac{1}{-\sin y} = \frac{-1}{\sqrt{1-\cos^2 y}} = \frac{-1}{\sqrt{1-x^2}} = \frac{1}{\sqrt{x^2-1}}

y := \arctan x \Leftrightarrow x = \tan y \Rightarrow \frac{dx}{dy} = \sec^2 y

\frac{dy}{dx} = \frac{1}{ \frac{dx}{dy}} = \frac{1}{ \sec^2 y} = \frac{1}{ 1+ \tan^2 y} = \frac{1}{1+x^2}

Note: In these derivations the “Pythagorean Identities” were used several times, and can be derived by dividing each term in the identity \sin^2 t + \cos^2 t = 1 by \sin^2 t \mbox{and} \cos^2 t respectively. In turn, that identity can be seen from the diagram in post #24.

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CONTRIBUTION & COMMENTS
\hrule
Just a quick reminder that this thread is still open to contribution from anyone and I would be more than happy to receive contributions and/or comments and suggestions for improvements/modifications.

However, could I request that in order to maintain some form of logical order and integrity that those wishing to contribute to this thread contact me via PM before contributing.

Thank you to those who have contributed already

 

[Hootenanny]

Differentiation of Series
In the previous section we discussed the derivatives if inverse functions, with the goal of discussing the derivatives of transcendental functions. However, before we can discuss transcendental functions, we must first examine the derivatives of series.

(1) Taylor Series
We have already met the Taylor series of some trigonometric functions in an early section. We shall now discuss "Taylor's Formula", which can be used to determine the [infinite] series representation for many functions. "Taylor's Theorem" defines the conditions for which a function must satisfy to be written as a Taylor series; however, it is not necessary to discuss the theorem here. We shall restrict ourselves to one-dimensional real-valued functions.

(1.1) Taylor's Formula
Given a function f:I\to\mathbb{R}, that is defined on some open interval I and which has derivatives of all orders at a point a\in I, then one may represent f(x) as,

f(x)=\sum_{n=1}^{\infty}\frac{f^{(n)}(a)}{n!}\left(x-a\right)^n = f(a)+f^\prime(a)\left(x-a\right)+\ldots​

(2) Term-wise Differentiation
Term-wise differentiation [and integration] is an extremely useful tool in analysis, which can be used to find the derivatives of functions which can be represented as [infinite] series. There is a theorem of advanced calculus which states that one can find the derivative of a [convergent] series by simply differentiating each term separately. We shall states this theorem here without proof.

(2.1) Theorem
Given an infinite series,

\sum_{n}^{\infty}f_n(x) = f_1(x)+f_2(x)+\ldots+f_n(x)+\ldots

that has continuous derivatives on some open interval I and converges [uniformally] to F(x), then if the series of differentiated terms,

\sum_{n}^{\infty}f^\prime_n(x) = f^\prime_1(x)+f^\prime_2(x)+\ldots+f^\prime_n(x)+\ldots

also converges uniformally on the interval I, then the original series can be differentiated term-wise such that,

\frac{d}{dx}F(x) = \sum_{n}^{\infty}f_n(x) = f^\prime_1(x)+f^\prime_2(x)+\ldots+f^\prime_n(x)+\ldots
​

Now we have all the necessary tools to examine the derivatives of transcendental functions, which is what we shall do in the following section.

 

[aggfx]

Would it be possible to make a PDF version of all of this for refrence like you did for integration?

 

[Hootenanny]

Would it be possible to make a PDF version of all of this for refrence like you did for integration?

Welcome to PF aggfx,

If I have some time later in the year I may make a PDF of the final version once I've finished it. There are still a few sections that I would like to add, so there's not much point in creating a PDF at this stage.

By the way, kurdt is the author of the excellent Intro to Integration tutorial.

 

[aggfx]

Thanks for the welcome. Keep up the good work! Sorry for the mix-up on the author. Either way, its great

 

[Sherazi]

Thank you very much for posting this summary! This is extremely helpful!