Math 355: Higher Derivatives and Derivatives of Trigonometric Functions

Page 1: Higher-Order Derivatives and Motion

Higher-Order Derivatives: Concept and Notations

When a function $f(x)$ is differentiated, the resulting function $f'(x)$ is called the first derivative. Because $f'(x)$ is itself a function of $x$, it can be differentiated again with respect to $x$. The derivative of the first derivative is called the second derivative, denoted $f''(x)$. Continuing this process yields higher-order derivatives.

Derivative Notations

• First Derivative:

  $f'(x) = \frac{df}{dx} = \frac{dy}{dx}$

• Second Derivative:

  $f''(x) = \frac{d^2f}{dx^2} = \frac{d^2y}{dx^2}$

• Third Derivative:

  $f'''(x) = \frac{d^3f}{dx^3} = \frac{d^3y}{dx^3}$

• Fourth Derivative:

  To avoid an unwieldy number of prime tick marks, roman numerals or parenthesized integer exponents are used for derivatives of order four and higher:

  $f^{(4)}(x) = \frac{d^4f}{dx^4} = \frac{d^4y}{dx^4}$

• $n$-th Derivative:

  For an arbitrary positive integer $n$:

  $f^{(n)}(x) = \frac{d^n f}{dx^n} = \frac{d^n y}{dx^n}$

Example: Higher Derivatives of a Power Function

Consider the power function $f(x) = x^n$, where $n$ is a positive integer.

Applying the power rule successively:

• First derivative:

  $f'(x) = n x^{n-1}$

• Second derivative:

  $f''(x) = n(n-1) x^{n-2}$

• Third derivative:

  $f'''(x) = n(n-1)(n-2) x^{n-3}$

• General $k$-th derivative ($k \le n$):

  $f^{(k)}(x) = \frac{n!}{(n-k)!} x^{n-k}$

• $n$-th derivative:

  $f^{(n)}(x) = n!$

• $(n+1)$-th derivative and beyond:

  $f^{(n+1)}(x) = 0$

Kinematic Applications: Position, Velocity, and Acceleration

In rectilinear motion along a coordinate axis, higher derivatives provide descriptions of motion:

• Position Function: $s(t)$ represents the position of an object along a coordinate line relative to the origin at time $t$.

• Velocity Function: The instantaneous rate of change of position with respect to time represents velocity $v(t)$, given by the first derivative:

  $v(t) = s'(t) = \frac{ds}{dt}$

• Acceleration Function: The instantaneous rate of change of velocity with respect to time represents acceleration $a(t)$, given by the second derivative of position (or first derivative of velocity):

  $a(t) = v'(t) = s''(t) = \frac{d^2s}{dt^2}$

Example Problem: Motion of a Particle

Find the velocity and acceleration functions in terms of $t$ for the position function:

$s(t) = t^4 - 2t$

Step 1: Differentiate to find Velocity

$v(t) = s'(t) = \frac{d}{dt}(t^4 - 2t)$

$v(t) = 4t^3 - 2$

Step 2: Differentiate again to find Acceleration

$a(t) = v'(t) = s''(t) = \frac{d}{dt}(4t^3 - 2)$

$a(t) = 12t^2$

Page 2: Projectile Motion Applications and Trigonometric Limits

Vertical Projectile Motion

Neglecting air resistance, the height $h$ (in meters) of an object launched vertically upward from the ground with an initial velocity $v_0$ is modeled by the kinematic formula:

$h(t) = v_0 t - \frac{1}{2} g t^2$

Where:

• $g = 9.81\,m/s^2$ is the acceleration due to Earth's gravity.

• $t$ is the elapsed time in seconds since release ($t \ge 0$).

• $v_0$ is the initial upward velocity at $t = 0$.

Substituting $g = 9.81\,m/s^2$ into the position function gives:

$h(t) = v_0 t - \frac{1}{2}(9.81) t^2 = v_0 t - 4.905 t^2$

Example Problem: Kinematic Analysis of a Projectile

Part a: Find the Velocity and Acceleration Functions

• Velocity Function:

  $v(t) = h'(t) = \frac{d}{dt}\left(v_0 t - \frac{1}{2} g t^2\right) = v_0 - g t$

  Substituting $g = 9.81\,m/s^2$:

  $v(t) = v_0 - 9.81 t$

• Acceleration Function:

  $a(t) = v'(t) = h''(t) = \frac{d}{dt}(v_0 - g t) = -g$

  $a(t) = -9.81\,m/s^2$

  The acceleration is constant and directed downward toward the Earth.

Part b: Find the Time When the Velocity is Equal to 0

The velocity reaches zero at the peak of the trajectory (maximum height):

$v(t) = 0$

$v_0 - g t = 0$

$g t = v_0$

$t = \frac{v_0}{g} = \frac{v_0}{9.81}$

Part c: Time Interval When the Object is Traveling Upward

An object moves upward when its velocity is strictly positive ($v(t) > 0$):

$v_0 - g t > 0$

$t < \frac{v_0}{g}$

Since the release occurs at $t = 0$, the time interval during which the object travels upward is:

$\left[0, \frac{v_0}{g}\right) = \left[0, \frac{v_0}{9.81}\right)$

Part d: Time Interval When the Object is Falling Downward

An object falls downward when its velocity is negative ($v(t) < 0$), continuing until it impacts the ground ($h(t) = 0$).

To find the impact time, solve for $h(t) = 0$:

$v_0 t - \frac{1}{2} g t^2 = 0$

$t\left(v_0 - \frac{1}{2} g t\right) = 0$

• $t = 0$ (time of launch)

• $t = \frac{2v_0}{g}$ (time of landing)

Therefore, the object is falling downward during the time interval:

$\left(\frac{v_0}{g}, \frac{2v_0}{g}\right] = \left(\frac{v_0}{9.81}, \frac{2v_0}{9.81}\right]$

Introduction to Derivatives of Trigonometric Functions

The fundamental rates of change for primary trigonometric functions are:

$\frac{d}{dx}(\sin(x)) = \cos(x)$

$\frac{d}{dx}(\cos(x)) = -\sin(x)$

Essential Limits and Identities for Derivations

The formal proofs of these derivative formulas rely on two fundamental limits and angle addition identities:

• Fundamental Trigonometric Limits:

  $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$

  $\lim_{x \to 0} \frac{1 - \cos(x)}{x} = 0$

• Trigonometric Angle Sum Formulas:

  $\cos(x + h) = \cos(x)\cos(h) - \sin(x)\sin(h)$

  $\sin(x + h) = \sin(x)\cos(h) + \cos(x)\sin(h)$

Limit-Definition Proof of $\frac{d}{dx}\sin(x)$

Using the definition of the derivative:

$\frac{d}{dx}\sin(x) = \lim_{h \to 0} \frac{\sin(x + h) - \sin(x)}{h}$

$\frac{d}{dx}\sin(x) = \lim_{h \to 0} \frac{\sin(x)\cos(h) + \cos(x)\sin(h) - \sin(x)}{h}$

$\frac{d}{dx}\sin(x) = \lim_{h \to 0} \left[ \sin(x) \left(\frac{\cos(h) - 1}{h}\right) + \cos(x) \left(\frac{\sin(h)}{h}\right) \right]$

$\frac{d}{dx}\sin(x) = \sin(x) \left( -\lim_{h \to 0} \frac{1 - \cos(h)}{h} \right) + \cos(x) \left(\lim_{h \to 0} \frac{\sin(h)}{h}\right)$

$\frac{d}{dx}\sin(x) = \sin(x)(0) + \cos(x)(1) = \cos(x)$

Page 3: Graphical Relationships and Sine/Cosine Derivatives

Graphical Relationship Between Functions and Their Derivatives

Successive Cyclic Differentiation

Differentiating sine and cosine repeatedly demonstrates a 4-step cyclic pattern:

1. $y = \sin(x)$

2. $y' = \cos(x)$

3. $y'' = -\sin(x)$

4. $y''' = -\cos(x)$

5. $y^{(4)} = \sin(x)$

Geometric Interpretation of the Graphs

• Derivative of $y = \sin(x)$ is $y = \cos(x)$:

  • At $x = 0$, the graph of $\sin(x)$ passes through the origin with a slope of $1$, corresponding to $\cos(0) = 1$.

  • At $x = \frac{\pi}{2}$, $\sin(x)$ reaches a crest (horizontal tangent line with slope $0$), corresponding to $\cos\left(\frac{\pi}{2}\right) = 0$.

  • At $x = \pi$, $\sin(x)$ decreases with a slope of $-1$, corresponding to $\cos(\pi) = -1$.

  • At $x = \frac{3\pi}{2}$, $\sin(x)$ reaches a trough (horizontal tangent line with slope $0$), corresponding to $\cos\left(\frac{3\pi}{2}\right) = 0$.

• Derivative of $y = \cos(x)$ is $y = -\sin(x)$:

  • At $x = 0$, $\cos(x)$ has a horizontal tangent line with slope $0$, matching $-\sin(0) = 0$.

  • At $x = \frac{\pi}{2}$, $\cos(x)$ crosses the horizontal axis with a downward slope of $-1$, matching $-\sin\left(\frac{\pi}{2}\right) = -1$.

  • At $x = \pi$, $\cos(x)$ reaches its minimum with a slope of $0$, matching $-\sin(\pi) = 0$.

  • At $x = \frac{3\pi}{2}$, $\cos(x)$ crosses the horizontal axis with an upward slope of $1$, matching $-\sin\left(\frac{3\pi}{2}\right) = -(-1) = 1$.

Worked Examples

Example 1: Differentiating Linear Combinations with Constants

Find the derivative of:

$f(x) = -4\cos(x) + \sin\left(\frac{\pi}{4}\right)$

Solution:

• Differentiate $-4\cos(x)$ with respect to $x$:

  $\frac{d}{dx}[-4\cos(x)] = -4(-\sin(x)) = 4\sin(x)$

• Notice that $\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$ is a constant containing no variable $x$. The derivative of any constant is $0$:

  $\frac{d}{dx}\left[\sin\left(\frac{\pi}{4}\right)\right] = 0$

• Combining the terms yields:

  $f'(x) = 4\sin(x) + 0 = 4\sin(x)$

Example 2: Sine with a Polynomial Argument (Chain Rule)

Find the derivative of:

$y = \sin(x^2)$

Solution:

Applying the Chain Rule, $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$, where the outer function is $f(u) = \sin(u)$ and inner function is $u = g(x) = x^2$:

$\frac{du}{dx} = 2x$

$\frac{dy}{dx} = \cos(x^2) \cdot \frac{d}{dx}(x^2)$

$\frac{dy}{dx} = 2x\cos(x^2)$

Example 3: Power of a Sine Function (Chain Rule)

Find the derivative of:

$y = \sin^2(x) = (\sin(x))^2$

Solution:

Applying the General Power Rule / Chain Rule, $\frac{d}{dx}[u^n] = n u^{n-1} \cdot \frac{du}{dx}$, where $u = \sin(x)$ and $n = 2$:

$\frac{dy}{dx} = 2(\sin(x))^{2-1} \cdot \frac{d}{dx}(\sin(x))$

$\frac{dy}{dx} = 2\sin(x)\cos(x)$

Using the double-angle identity $\sin(2x) = 2\sin(x)\cos(x)$, this can also be expressed as:

$\frac{dy}{dx} = \sin(2x)$

Page 4: Derivatives of Other Trigonometric Functions and Complete Summary

Advanced Derivative Calculations

Example 4: Mixed Product and Chain Rule Expression

Find the derivative of:

$y = x^2 \sin(3x) - \cos(5x)$

Solution:

Separate the expression into two distinct parts: $y = u(x) - v(x)$, where $u(x) = x^2 \sin(3x)$ and $v(x) = \cos(5x)$.

• Differentiating $u(x) = x^2 \sin(3x)$ via the Product Rule:

  $\frac{d}{dx}[x^2 \sin(3x)] = \frac{d}{dx}[x^2] \cdot \sin(3x) + x^2 \cdot \frac{d}{dx}[\sin(3x)]$

  By the Chain Rule, $\frac{d}{dx}[\sin(3x)] = \cos(3x) \cdot 3 = 3\cos(3x)$.

  $\frac{d}{dx}[u(x)] = 2x\sin(3x) + 3x^2\cos(3x)$

• Differentiating $v(x) = \cos(5x)$ via the Chain Rule:

  $\frac{d}{dx}[\cos(5x)] = -\sin(5x) \cdot 5 = -5\sin(5x)$

• Combining all components:

  $\frac{dy}{dx} = \left(2x\sin(3x) + 3x^2\cos(3x)\right) - (-5\sin(5x))$

  $\frac{dy}{dx} = 2x\sin(3x) + 3x^2\cos(3x) + 5\sin(5x)$

Example 5: Derivation of the Derivative of $\tan(x)$

Find the derivative of $y = \tan(x)$.

Solution:

Rewrite $\tan(x)$ using the quotient definition $\tan(x) = \frac{\sin(x)}{\cos(x)}$ and apply the Quotient Rule:

$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$

$\frac{d}{dx}\tan(x) = \frac{\frac{d}{dx}[\sin(x)] \cdot \cos(x) - \sin(x) \cdot \frac{d}{dx}[\cos(x)]}{\cos^2(x)}$

$\frac{d}{dx}\tan(x) = \frac{\cos(x)\cos(x) - \sin(x)(-\sin(x))}{\cos^2(x)}$

$\frac{d}{dx}\tan(x) = \frac{\cos^2(x) + \sin^2(x)}{\cos^2(x)}$

Applying the Pythagorean identity $\cos^2(x) + \sin^2(x) = 1$:

$\frac{d}{dx}\tan(x) = \frac{1}{\cos^2(x)} = \left(\frac{1}{\cos(x)}\right)^2 = \sec^2(x)$

Example 6: Derivation of the Derivative of $\sec(x)$

Find the derivative of $y = \sec(x)$.

Solution:

Rewrite $\sec(x)$ as the reciprocal of cosine: $\sec(x) = (\cos(x))^{-1}$.

Apply the Chain Rule:

$\frac{d}{dx}\sec(x) = -1(\cos(x))^{-2} \cdot \frac{d}{dx}[\cos(x)]$

$\frac{d}{dx}\sec(x) = -\frac{1}{\cos^2(x)} \cdot (-\sin(x))$

$\frac{d}{dx}\sec(x) = \frac{\sin(x)}{\cos^2(x)}$

Factor the denominator to express using basic reciprocal functions:

$\frac{d}{dx}\sec(x) = \frac{1}{\cos(x)} \cdot \frac{\sin(x)}{\cos(x)} = \sec(x)\tan(x)$

Summary of Basic Trigonometric Derivatives

Complete List of Derivatives

$\frac{d}{dx}\sin(x) = \cos(x)$

$\frac{d}{dx}\cos(x) = -\sin(x)$

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

$\frac{d}{dx}\cot(x) = -\csc^2(x)$

$\frac{d}{dx}\sec(x) = \sec(x) \cdot \tan(x)$

$\frac{d}{dx}\csc(x) = -\csc(x) \cdot \cot(x)$

Structural Patterns to Note

• Negative Signs on "Co-" Functions: The derivatives of all three co-functions ($\cos(x)$, $\cot(x)$, and $\csc(x)$) contain a negative sign.

• Pairs of Functions:

  • Tangent maps to secant squared ($\sec^2(x)$), whereas cotangent maps to negative cosecant squared ($-\csc^2(x)$).

  • Secant differentiates into the product of itself and tangent ($\sec(x)\tan(x)$), whereas cosecant differentiates into the negative product of itself and cotangent ($-\csc(x)\cot(x)$).