Calculus Engage

Module 2: Trigonometric and Inverse Trigonometric Functions

This module presents a review of the fundamentals of trigonometric and inverse trigonometric functions, including their properties and graphs.

5 Lessons

13hrs 30mins

26 students enrolled

Calculus 1

Love Quinn

Instructor

5 (26)

Module Content

Disclaimer

All images such as logos, photos, etc., used in this module are the property of their respective copyright owners and are used here for educational purposes only.

Introduction

This module presents a review of the fundamentals of trigonometric and inverse trigonometric functions, including their properties and graphs. Like algebraic functions, trigonometric functions also have standard derivative formulas, which will be covered in this chapter.

Intended Learning Outcomes

Recognize trigonometric and inverse trigonometric functions along with their corresponding graphs.
Understand the fundamental derivative formulas for trigonometric and inverse trigonometric functions.
Compute the derivatives of expressions that include trigonometric and inverse trigonometric functions.

Lesson Overview

5 Lesssons

Lecture: Introduction and discussion on the basic trigonometric functions: a review.
Examples 1-6

Example graphs 1 - 6

Lecture: Introduction and discussion on the derivative of trigonometric functions.
Sample Problem 1-6: Differentiate the function.

Lecture: Introduction and discussion on the inverse trigonometric functions and its graphs.

Lecture: Introduction and discussion on the derivatives of inverse trigonometric functions.
Sample Problem 1-5: Find the derivative.

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$$\begin{aligned} (\text{hypotenuse})^2 &= (\text{opposite side})^2 + (\text{adjacent side})^2 \end{aligned}$$

“For any right triangle, the square of the longest side is equal to the sum of the squares of the other two sides.”

$$\begin{aligned} \sin^2 A + \cos^2 A &= 1 \\ \tan^2 A + 1 &= \sec^2 A \\ \cot^2 A + 1 &= \csc^2 A \end{aligned}$$

$$\begin{aligned} \sin(x \pm y) &= \sin x \cos y \pm \cos x \sin y \\ \cos(x \pm y) &= \cos x \cos y \mp \sin x \sin y \\ \tan(x \pm y) &= \frac{\tan x \pm \tan y}{1 \mp \tan x \tan y} \end{aligned}$$

$$\begin{aligned} \sin(2x) &= 2 \sin x \cos x, \\ \cos(2x) &= \cos^2 x - \sin^2 x \\ &= 1 - 2\sin^2 x \\ &= 2\cos^2 x - 1, \\ \tan(2x) &= \frac{2 \tan x}{1 - \tan^2 x}. \end{aligned}$$

$$\begin{aligned} \sin^2 x &= \frac{1 - \cos 2x}{2},\ \\cos^2 x &= \frac{1 + \cos 2x}{2},\ \\tan^2 x &= \frac{1 - \cos 2x}{1 + \cos 2x}. \end{aligned}$$

$$\begin{aligned} \sin x \cos y &= \frac{\sin(x+y) + \sin(x-y)}{2},\ \\sin x \sin y &= \frac{\cos(x-y) - \cos(x+y)}{2},\ \\cos x \cos y &= \frac{\cos(x+y) + \cos(x-y)}{2}. \end{aligned}$$

Sample Problem 1-4: Derive the functions.

Derive the function $ y = 3x^{2} + 5x - 2 $

By the general power formula:

\[ \frac{dy}{dx} = y' = 6x + 5 \]

Derive the function $ y = \frac{x^2 + 1}{x^3} $

By the derivative of a quotient:

\[ \begin{aligned} y' &= \frac{x^3 (2x \, dx) - (x^2 + 1)(3x^2 \, dx)}{(x^3)^2} \\ y' &= \frac{2x^4 \, dx - 3x^4 \, dx - 3x^2 \, dx}{x^6} \\ y' &= \frac{(-x^4 - 3x^2) dx }{x^6} \\ y' &= - \frac{x^2 + 3}{x^4} \ \end{aligned} \]

Derive the function $ y = \frac{1}{x^2} + \frac{2}{x^3} $

By the derivative of a quotient of a constant and a function:

\[ \begin{aligned} y' &= - \frac{2x \, dx}{x^4} - \frac{2(3x^2) \, dx}{x^6} \\ y' &= - \frac{2}{x^3} - \frac{6}{x^4} \end{aligned} \]

Derive the function $ y = (t^2 + 5)(t^2 - t + 3) $.

By the derivative of a product:

\[ \begin{aligned} y &= (t^2 + 5)(t^2 - t + 3) \\ dy &= (t^2 + 5)(2t - 1) \, dt + (t^2 - t + 3)(2t) \, dt \\ \frac{dy}{dt} &= 2t^3 - t^2 + 10t - 5 + 2t^3 - 2t^2 + 6t \\ dy &= 4t^3 - 3t^2 + 16t - 5 \, dt \\ \end{aligned} \]

Given the function $ y = \sin u $, the derivative of the function is,

\[ \frac{d}{dx} \sin u = \cos u \frac{du}{dx} \]

Proof:

\[ \begin{align*} y & = \sin u \\ y + \Delta y & = \sin(u + \Delta u) \\ y + \Delta y & = \sin u \cos \Delta u + \cos u \sin \Delta u \\ \Delta y & = \sin u \cos \Delta u + \cos u \sin \Delta u - \sin u \\ \Delta y & = \sin u (\cos \Delta u - 1) + \cos u \sin \Delta u \\ \Delta y & = \cos u \sin \Delta u - \sin u (1 - \cos \Delta u) \\ \Delta y & = \cos u \sin \Delta u - 2 \sin u \sin^2\left(\frac{1}{2} \Delta u\right) \\ \frac{\Delta y}{\Delta u} & = \cos u \left( \frac{\sin \Delta u}{\Delta u} \right) - 2 \sin u \left[ \frac{\sin^2\left(\frac{1}{2} \Delta u\right)}{\Delta u} \right] \\ \frac{\Delta y}{\Delta u} & = \cos u \left( \frac{\sin \Delta u}{\Delta u} \right) - \sin u \left[ \frac{\sin\left(\frac{1}{2} \Delta u\right)}{\frac{1}{2} \Delta u} \right] \left[\sin\left(\frac{1}{2} \Delta u\right) \right] \\ \frac{dv}{du} & = \lim_{\Delta u \to 0} \frac{\Delta y}{\Delta u} \\ & = \lim_{\Delta u \to 0} \left\{ \cos u \left( \frac{\sin \Delta u}{\Delta u} \right) - \sin u \left[ \frac{\sin\left(\frac{1}{2} \Delta u\right)}{\frac{1}{2} \Delta u} \right] \left[ \sin\left(\frac{1}{2} \Delta u\right) \right] \right\} \\ \frac{dy}{du} & = \cos u \\ \frac{dy}{dx} & = \cos u \frac{du}{dx} \end{align*} \]

Given the function $ y = \cos u $, the derivative of the function is,

\[ \frac{d}{dx} \cos u = -\sin u \frac{du}{dx} \]

Proof:

\[ \begin{align*} y & = \cos u \\ y + \Delta y & = \cos(u + \Delta u) \\ y + \Delta y & = \cos u \cos \Delta u - \sin u \sin \Delta u \\ \Delta y & = \cos u \cos \Delta u - \sin u \sin \Delta u - \cos u \\ \Delta y & = \cos u (\cos \Delta u - 1) - \sin u \sin \Delta u \\ \Delta y & = -\sin u \sin \Delta u - \cos u (1 - \cos \Delta u) \\ \Delta y & = -\sin u \sin \Delta u - 2 \cos u \sin^2\left(\frac{1}{2} \Delta u\right) \\ \frac{\Delta y}{\Delta u} & = -\sin u \left( \frac{\sin \Delta u}{\Delta u} \right) - 2 \cos u \left[ \frac{\sin^2\left(\frac{1}{2} \Delta u\right)}{\Delta u} \right] \\ \frac{\Delta y}{\Delta u} & = -\sin u \left( \frac{\sin \Delta u}{\Delta u} \right) - \cos u \left[ \frac{\sin\left(\frac{1}{2} \Delta u\right)}{\frac{1}{2} \Delta u} \right] \left[\sin\left(\frac{1}{2} \Delta u\right) \right] \\ \frac{d y}{d u} & = \lim_{\Delta u \to 0} \frac{\Delta y}{\Delta u} \\ & = \lim_{\Delta u \to 0} \left\{ -\sin u \left( \frac{\sin \Delta u}{\Delta u} \right) - \cos u \left[ \frac{\sin\left(\frac{1}{2} \Delta u\right)}{\frac{1}{2} \Delta u} \right] \left[ \sin\left(\frac{1}{2} \Delta u\right) \right] \right\} \\ \frac{d y}{d u} & = -\sin u \\ \frac{d y}{d x} & = -\sin u \frac{d u}{d x} \end{align*} \]

Given the function $ y = \tan u $, the derivative of the function is,

\[ \frac{d}{dx} \tan u = \sec^2 u \frac{du}{dx} \]

Proof:

\[ \begin{align*} y & = \tan u \\ y & = \frac{\sin u}{\cos u} \end{align*} \]

By the derivative of a quotient, \[ \frac{dy}{du} = \frac{ \cos u \frac{d}{du}(\sin u) - \sin u \frac{d}{du}(\cos u) }{ \cos^2 u } \] \[ \frac{dy}{du} = \frac{ \cos u \cos u + \sin u \sin u }{ \cos^2 u } \]

From the trigonometric identity, \[ \sin^2 u + \cos^2 u = 1 \] \[ \frac{dy}{du} = \frac{1}{\cos^2 u} = \sec^2 u \] \[ \frac{dy}{dx} = \sec^2 u \frac{du}{dx} \]

Given the function $ y = \cot u $, the derivative of the function is,

\[ \frac{d}{dx} \cot u = -\csc^2 u \frac{du}{dx} \]

Proof:

\[ \begin{align*} y & = \cot u \\ y & = \frac{\cos u}{\sin u} \\[1.5em] \frac{dy}{du} & = \frac{ \sin u \cdot (-\sin u) - \cos u \cdot \cos u }{ \sin^2 u } \\ \frac{dy}{du} & = \frac{ -\sin^2 u - \cos^2 u }{ \sin^2 u } \\ \frac{dy}{du} & = -\frac{ \sin^2 u + \cos^2 u }{ \sin^2 u } \\ \frac{dy}{du} & = -\frac{1}{\sin^2 u} = -\csc^2 u \\[1.5em] \frac{dy}{dx} & = -\csc^2 u \left( \frac{du}{dx} \right) \end{align*} \]

Given the function $ y = \sec u $, the derivative of the function is,

\[ \frac{d}{dx} \sec u = \sec u \tan u \frac{du}{dx} \]

Proof:

\[ \begin{align*} y & = \sec u \\ y & = \frac{1}{\cos u} \\[1.5em] \frac{dy}{du} & = \frac{ -\sin u }{ \cos^2 u } \\[1.5em] \frac{dy}{du} & = \frac{ \sin u }{ \cos^2 u } = \sec u \tan u \\[1.5em] \frac{dy}{dx} & = \sec u \tan u \frac{du}{dx} \end{align*} \]

Given the function $ y = \csc u $, the derivative of the function is,

\[ \frac{d}{dx} \csc u = -\csc u \cot u \frac{du}{dx} \]

Proof:

\[ \begin{align*} y & = \csc u \\ y & = \frac{1}{\sin u} \\[1.5em] \frac{dy}{du} & = \frac{ -\cos u }{ \sin^2 u } = -\csc u \cot u \\[1.5em] \frac{dy}{dx} & = -\csc u \cot u \frac{du}{dx} \end{align*} \]

\[ \frac{d}{dx}(\sin u) = \cos u \frac{du}{dx} \] \[ \frac{d}{dx}(\cot u) = -\csc^2 u \frac{du}{dx} \]

\[ \frac{d}{dx}(\cos u) = -\sin u \frac{du}{dx} \] \[ \frac{d}{dx}(\sec u) = \sec u \tan u \frac{du}{dx} \]

\[ \frac{d}{dx}(\tan u) = \sec^2 u \frac{du}{dx} \] \[ \frac{d}{dx}(\csc u) = -\cot u \csc u \frac{du}{dx} \]

Differentiate the function $ y = \sin 4x \cos x $.

\[ \begin{align*} y & = \sin 4x \cos x \\ \frac{dy}{dx} & = \sin 4x (-\sin x) + \cos x [\cos 4x \cdot 4] \\ \frac{dy}{dx} & = 4 \cos 4x \cos x - \sin 4x \sin x \end{align*} \]

Differentiate the function $ y = x^2 \cot x $.

\[ \begin{align*} y & = x^2 \cot x \\ \frac{dy}{dx} & = x^2 (-\csc^2 x) + \cot x (2x) \\ \frac{dy}{dx} & = -x^2 \csc^2 x + 2x \cot x \\ \frac{dy}{dx} & = y = x (2\cot x - x \csc^2 x) \end{align*} \]

Find the derivative $ xy = (1 + \cos^3 x)^{1/2} $

\[ \begin{align*} xy & = \left( 1 + \cos^3 x \right)^{1/2} \\[1.5em] \text{By implicit differentiation,} \\[0.5em] x \, dy + y \, dx & = \frac{3 \cos^2 x (-\sin x \, dx)}{2 \sqrt{1 + \cos^3 x}} \\[1.5em] x \, dy & = \frac{ -3 \cos^2 x \sin x }{ 2 \sqrt{1 + \cos^3 x} } dx - y \, dx \\[1.5em] \frac{dy}{dx} & = \frac{ -3 \cos^2 x \sin x }{ 2x \sqrt{1 + \cos^3 x} } - \frac{y}{x} \end{align*} \]

Find the first five derivatives of $ y = \sin ax + \cos bx $ where $ a $ and $ b $ are constants.

\[ \begin{align*} y & = \sin a x + \cos b x \\ y' & = a \cos a x - b \sin b x \\ y'' & = -a^2 \sin a x - b^2 \cos b x \\ y''' & = -a^3 \cos a x + b^3 \sin b x \\ y^{(4)} & = a^4 \sin a x + b^4 \cos b x \\ y^{(5)} & = a^5 \cos a x - b^5 \sin b x \end{align*} \]

Find the derivative of $ y = 2 \tan(x/2) - x $

By chain rule and derivative of a constant: \[ \begin{align*} y' & = 2 \cdot \sec^2 (x/2) \cdot (1/2) - 1 \\ & = \sec^2 (x/2) - 1 \end{align*} \]

Find the derivative $ y = \frac{2\cot x}{1 + \cot^2 x} $

By quotient rule:
Let $ u = 2 \cot x $, $ du = -2 \csc^2 x $
$ v = 1 + \cot^2 x $, $ dv = 2 \cot x (-\csc^2 x) $

Formula: $ y' = \frac{v du - u dv}{v^2} $

\[ \begin{align*} y' & = \frac{(1 + \cot^2 x)(-2\csc^2 x) - (2\cot x)\left[2\cot x(-\csc^2 x)\right]}{(1 + \cot^2 x)^2} \\[1.2em] & = \frac{-2\csc^2 x (1 + \cot^2 x) + 4 \cot^2 x \csc^2 x}{(1 + \cot^2 x)^2} \\[1.2em] & = \frac{2\csc^2 x \left[-(1 + \cot^2 x) + 2\cot^2 x \right]}{(1 + \cot^2 x)^2} \\[1.2em] & = \frac{2\csc^2 x \left[-1 - \cot^2 x + 2\cot^2 x \right]}{(1 + \cot^2 x)^2} \\[1.2em] & = \frac{2\csc^2 x \left[-1 + \cot^2 x \right]}{(1 + \cot^2 x)^2} \\[1.2em] & = \frac{2(\cot^2 x - 1)}{\sin^2 x (1 + \cot^2 x)^2} \\[1.2em] \end{align*} \]

But $ 1 + \cot^2 x = \csc^2 x $, so: \[ \begin{align*} y' & = \frac{2(\cot^2 x - 1)}{\csc^4 x} \\[1.2em] & = 2 \sin^4 x (\cot^2 x - 1) \end{align*} \]

But $ \cot^2 x = \frac{\cos^2 x}{\sin^2 x} $, so: \[ \begin{align*} y' & = 2 \sin^2 x \left(\cos^2 x - \sin^2 x \right) \\[1.2em] & = 2 \sin^2 x \cos 2x \end{align*} \]

So the derivative is: \[ y' = 2 \cos 2x \]

Given the trigonometric function $ \sin y = x $, the inverse of the function is $ y = \sin^{-1} x $. Some inverse trigonometric functions are as follows:

$ y = \cos^{-1} x $ if $ \cos y = x $
$ y = \tan^{-1} x $ if $ \tan y = x $
$ y = \cot^{-1} x $ if $ \cot y = x $

$ y = \sec^{-1} x $ if $ \sec y = x $
$ y = \csc^{-1} x $ if $ \csc y = x $

Derivative of inverse sine function

Given the function $ y = \sin^{-1} u $, since $ \sin y = u $, the derivative of the function is:

\[ \frac{d}{dx} \sin^{-1} u = \frac{\frac{du}{dx}}{\sqrt{1-u^2}} \]

Proof:

Derive the function $ \sin y = u $:

\[ \cos y \frac{dy}{dx} = \frac{du}{dx} \]

\[ \frac{dy}{dx} = \frac{ 1 }{\cos y} \]

From the trigonometric identity:

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

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

Since $ u = \sin y $, thus:

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

Therefore:

\[ \frac{dy}{dx} = \frac{1}{\sqrt{1-u^2}} \]

\[ \frac{dy}{dx} = \frac{\frac{du}{dx}}{\sqrt{1-u^2}} \]

Given the function:

\[ y = \cos^{-1} u \]

Since:

\[ \cos y = u \]

The derivative of the function is:

\[ \frac{d}{dx} \cos^{-1} u = - \frac{du/dx}{\sqrt{1-u^2}} \]

Proof:

Derive the function:

\[ - \sin y \frac{dy}{dx} = \frac{du}{dx} \]

Solving for $ \frac{dy}{dx} $:

\[ \frac{dy}{dx} = - \frac{1}{\sin y} \]

Using the trigonometric identity:

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

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

Since $ u = \cos y $, then:

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

Thus,

\[ \frac{dy}{dx} = - \frac{1}{\sqrt{1-u^2}} \]

Which simplifies to:

\[ \frac{dy}{dx} = - \frac{du/dx}{\sqrt{1-u^2}} \]

Given the function:

\[ y = \tan^{-1} u \]

Since:

\[ \tan y = u \]

The derivative of the function is:

\[ \frac{d}{dx} \tan^{-1} u = \frac{\frac{du}{dx}}{1+u^2} \]

Proof:

Derive the function:

\[ \sec^2 y \frac{dy}{dx} = \frac{du}{dx} \]

Solving for $ \frac{dy}{dx} $:

\[ \frac{dy}{dx} = \frac{1}{\sec^2 y} \]

Using the trigonometric identity:

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

Since $ u = \tan y $, then:

\[ \sec^2 y = 1 + u^2 \]

Thus,

\[ \frac{dy}{dx} = \frac{1}{1+u^2} \]

Which simplifies to:

\[ \frac{dy}{dx} = \frac{\frac{du}{dx}}{1+u^2} \]

Given the function:

\[ y = \cot^{-1} u \]

Since:

\[ \cot y = u \]

The derivative of the function is:

\[ \frac{d}{dx} \cot^{-1} u = - \frac{\frac{du}{dx}}{1 + u^2} \]

Proof:

Derive the function:

\[ - \csc^2 y \frac{dy}{dx} = \frac{du}{dx} \]

Solving for $ \frac{dy}{dx} $:

\[ \frac{dy}{dx} = - \frac{1}{\csc^2 y} \]

Using the trigonometric identity:

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

Since $ u = \cot y $, then:

\[ \csc^2 y = 1 + u^2 \]

Thus,

\[ \frac{dy}{dx} = - \frac{1}{1 + u^2} \]

Which simplifies to:

\[ \frac{dy}{dx} = - \frac{\frac{du}{dx}}{1 + u^2} \]

Given the function:

\[ y = \sec^{-1} u \]

Since:

\[ \sec y = u \]

The derivative of the function is:

\[ \frac{d}{dx} \sec^{-1} u = \frac{\frac{du}{dx}}{u\sqrt{u^2-1}} \]

Proof:

Derive the function:

\[ \sec y \tan y \frac{dy}{dx} = \frac{du}{dx} \]

Solving for $ \frac{dy}{dx} $:

\[ \frac{dy}{dx} = \frac{1}{\sec y \tan y} \]

Using the trigonometric identity:

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

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

Since $ u = \sec y $, then:

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

Thus,

\[ \frac{dy}{dx} = \frac{1}{u\sqrt{u^2 - 1}} \]

Which simplifies to:

\[ \frac{dy}{dx} = \frac{\frac{du}{dx}}{u\sqrt{u^2 - 1}} \]

Given the function:

\[ y = \csc^{-1} u \]

Since:

\[ \csc y = u \]

The derivative of the function is:

\[ \frac{d}{dx} \csc^{-1} u = -\frac{du}{dx} \frac{1}{u\sqrt{u^2 - 1}} \]

Proof:

Derive the function:

\[ -\csc y \cot y \frac{dy}{dx} = \frac{du}{dx} \]

Solving for $ \frac{dy}{dx} $:

\[ \frac{dy}{dx} = -\frac{1}{\csc y \cot y} \]

Using the trigonometric identity:

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

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

Since $ u = \csc y $, then:

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

Thus,

\[ \frac{dy}{dx} = -\frac{1}{u\sqrt{u^2 - 1}} \]

Which simplifies to:

\[ \frac{dy}{dx} = -\frac{du}{dx} \frac{1}{u\sqrt{u^2 - 1}} \]

\[ \frac{d}{dx} (\sin^{-1} u) = \frac{du}{dx} \cdot \frac{1}{\sqrt{1 - u^2}} \]

\[ \frac{d}{dx} (\cos^{-1} u) = -\frac{du}{dx} \cdot \frac{1}{\sqrt{1 - u^2}} \]

\[ \frac{d}{dx} (\tan^{-1} u) = \frac{du}{dx} \cdot \frac{1}{1 + u^2} \]

\[ \frac{d}{dx} (\cot^{-1} u) = -\frac{du}{dx} \cdot \frac{1}{1 + u^2} \]

\[ \frac{d}{dx} (\sec^{-1} u) = \frac{du}{dx} \cdot \frac{1}{|u| \sqrt{u^2 - 1}} \]

\[ \frac{d}{dx} (\csc^{-1} u) = -\frac{du}{dx} \cdot \frac{1}{|u| \sqrt{u^2 - 1}} \]

Differentiate the function:

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

From:

\[ \frac{d}{dx} \tan^{-1} u = \frac{\frac{du}{dx}}{1 + u^2} \]

We get:

\[ \begin{align*} y' & = \frac{2(x + 2)}{1 + [(x + 2)^2]^2} \\ y' & = \frac{2(x + 2)}{1 + (x + 2)^4} \end{align*} \]

Find the \[ \frac{dx}{dt} \] of the equation:

\[ xt^2 = x - \cos^{-1}(t^2) \]

Differentiate both sides with respect to $ t $:

\[ \begin{align*} \frac{d}{dt}(xt^2) & = \frac{d}{dt}\left(x - \cos^{-1}(t^2)\right) \\ x \cdot \frac{d}{dt}(t^2) + t^2 \cdot \frac{dx}{dt} & = \frac{dx}{dt} - \left( \frac{d}{dt} \cos^{-1}(t^2) \right) \\ 2tx + t^2 \cdot \frac{dx}{dt} & = \frac{dx}{dt} - \left( \frac{-2t}{\sqrt{1 - (t^2)^2}} \right) \\ t^2 \cdot \frac{dx}{dt} - \frac{dx}{dt} & = \frac{2t}{\sqrt{1 - t^4}} - 2tx \\ (t^2 - 1) \cdot \frac{dx}{dt} & = \frac{2t}{\sqrt{1 - t^4}} - 2tx \\ \frac{dx}{dt} & = \frac{2t}{t^2 - 1} \left( \frac{1}{\sqrt{1 - t^4}} - x \right) \end{align*} \]

Find the second derivative of:

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

First derivative:

\[ \begin{align*} y' & = \frac{d}{dx} \left( x^2 + \cot^{-1} x \right) \\ & = 2x - \frac{1}{1 + x^2} \end{align*} \]

Second derivative:

\[ \begin{align*} y'' & = \frac{d}{dx} \left( 2x - \frac{1}{1 + x^2} \right) \\ & = 2 + \frac{2x}{(1 + x^2)^2} \end{align*} \]

Find the derivative of:

\[ x = \sec^{-1}(5t) + \csc^{-1}(5t) \]

Derivative:

\[ \begin{align*} x' & = \frac{1}{5t \sqrt{25t^2 - 1}} (5) - \frac{1}{5t \sqrt{25t^2 - 1}} (5) \\ x' & = 0 \end{align*} \]

Find the derivative of:

\[ y = x^2 \tan^{-1}(3x) \]

Using the product rule: $ y' = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx} $

\[ \begin{align*} u & = x^2 \\ v & = \tan^{-1}(3x) \\ \frac{du}{dx} & = 2x \\ \frac{dv}{dx} & = \frac{3}{1 + 9x^2} \end{align*} \]

Therefore:

\[ \begin{align*} y' & = x^2 \cdot \frac{3}{1 + 9x^2} + 2x \cdot \tan^{-1}(3x) \\ & = \frac{3x^2}{1 + 9x^2} + 2x \tan^{-1}(3x) \end{align*} \]

Factored form:

\[ y' = x \left( \frac{3x}{1 + 9x^2} + 2 \tan^{-1}(3x) \right) \]

\[ \frac{d}{dx}(\sin u) = \cos u \frac{du}{dx} \]	\[ \frac{d}{dx}(\cot u) = -\csc^2 u \frac{du}{dx} \]
\[ \frac{d}{dx}(\cos u) = -\sin u \frac{du}{dx} \]	\[ \frac{d}{dx}(\sec u) = \sec u \tan u \frac{du}{dx} \]
\[ \frac{d}{dx}(\tan u) = \sec^2 u \frac{du}{dx} \]	\[ \frac{d}{dx}(\csc u) = -\cot u \csc u \frac{du}{dx} \]

Module 2: Trigonometric and Inverse Trigonometric Functions

Love Quinn

Module Content

Disclaimer

Introduction

Intended Learning Outcomes

Lesson Overview

5 Lesssons

2.1 Trigonometric Functions: A review

2.2 Graphs of Trigonometric Functions

2.3 Derivative of Trigonometric Functions

2.4 Inverse Trigonometric Functions

2.5 Derivatives of Inverse Trigonometric Functions

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