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Lecture 6: Basic Derivative Rules

GOAL: Compute derivatives without relying on the definition.

Summary:

\[ \begin{array}{l|l} \dfrac{d}{dx}(x^p) = px^{p-1} & \dfrac{d}{dx}(e^x) = e^x \\[2ex] \dfrac{d}{dx}(\sin x) = \cos x & \dfrac{d}{dx}(\cos x) = -\sin x \\[2ex] \dfrac{d}{dx}(f(x) \pm g(x)) = \dfrac{df}{dx} + \dfrac{dg}{dx} & \dfrac{d}{dx}(k \cdot f(x)) = k\dfrac{df}{dx} \end{array} \]

Polynomials and Power Functions

We compute the derivative of a constant \( c \) as well as \(x\):

\[ \begin{gathered} (c)' = \lim_{h \to 0} \frac{c - c}{h} = 0 \\[0.5em] (x)' = \lim_{h \to 0} \frac{(x+h) - x}{h} = 1 \end{gathered} \]
Functions, degrees, and derivatives
Function Degree Derivative
Constant \( 0 \) \( 0 \)
\( x \) \( 1 \) \( 1 = 1 \cdot x^{1-1} \)
\( x^2 \) \( 2 \) \( 2x = 2x^{2-1} \)
\( x^3 \) \( 3 \) \( 3x^2 = 3x^{3-1} \)
\( x^4 \) \( 4 \) \( 4x^3 = 4x^{4-1} \)

Computing the derivatives of \(x^p\) for small integer values of \(p\), we see a pattern. This pattern is true for any constant in the exponent.

Theorem (Power Rule)

Let \( p \) be any real number. Then

\[ \frac{d}{dx}\left(x^p\right) = p \cdot x^{(p)-1} = p x^{p-1} \]

Why? We will show in steps over several lectures.

Example 1

  1. \( p = 10 \)

    \[ \frac{d}{dx}\left(x^{10}\right) = 10 \cdot x^{(10)-1} = 10x^9 \]

  2. \( p = -2 \)

    \[ \frac{d}{dx}\left(\frac{1}{x^2}\right) = \frac{d}{dx}\left(x^{-2}\right) = -2 \cdot x^{(-2)-1} = -2x^{-3} = \frac{-2}{x^3} \]

  1. \( p = \frac{1}{3} \)

    \[ \begin{gathered} \frac{d}{dx}\left(\sqrt[3]{x}\right) = \frac{d}{dx}\left(x^{\frac{1}{3}}\right) = \frac{1}{3} \cdot x^{\frac{1}{3}-1} = \frac{1}{3} x^{-\frac{2}{3}} = \frac{1}{3\sqrt[3]{x^2}} \end{gathered}\]

Here, we needed \(x^{\frac{p}{q}} = \sqrt[q]{x^p} = \left(\sqrt[q]{x}\right)^p\) for integers \(p\) and \(q\neq 0\).

Question: How can we differentiate polynomials like \( x^2 - 2x + 3 \)?

Theorem [Linearity]

Let \( f \) and \( g \) be differentiable functions

  1. Sum/Difference Rule

    \[ [f(x) \pm g(x)]' = f'(x) \pm g'(x) \]

  2. Constant-Multiple Rule

    \[ [k \cdot f(x)]' = k [f(x)]' \quad \text{for constant } k \]

Example 2

\[ \begin{aligned} [x^2 - 2x + 3]' &= [x^2]' - [2x]' + [3]' \\ &= [x^{2}]' - 2[x]' + [3]' \\ &= 2 \cdot x^{2-1} - 2 \cdot 1x^{1-1} + 0 \\ &= 2x - 2 \end{aligned} \]

Example 3 (HW6, #8)

\begin{align*} \left[\frac{(2x)^{\pi} -x^{3.5}}{\sqrt{x}}\right]' =&~ \left[\frac{2^{\pi} x^{\pi} - x^{3.5}}{x^{0.5}}\right]'\\ =&~ \left[\frac{2^{\pi}x^{\pi}}{x^{0.5}} - \frac{x^{3.5}}{x^{0.5}}\right]' \\[0.5em] =&~ \left[2^{\pi} x^{\pi - 0.5} - x^{3}\right]'\\ =&~2^{\pi}\left[x^{\underbrace{\pi - 0.5}_{p}}\right]' - \left[x^3\right]' \\ =&~ 2^{\pi} \cdot (\pi - 0.5) x^{\pi - 0.5 - 1} - 3 \cdot x^{3-1} \\ =&~ 2^{\pi}(\pi - 0.5) x^{\pi - 1.5} - 3x^2 \end{align*}

Sine and Cosine

Theorem

  1. \[ \frac{d}{dx}(\sin x) = \cos x \]
  2. \[ \frac{d}{dx}(\cos x) = - \sin x\]

Don't forget the minus sign in the derivative of cosine.

Why?

  1. See the Textbook.

  2. Recall \( \cos(a+b) = \cos a \cos b - \sin a \sin b \)

    \[ \begin{aligned} \frac{d}{dx}(\cos x) &= \lim_{h \to 0} \frac{\cos(x+h) - \cos x}{h} \\ &= \lim_{h \to 0} \frac{\cos x \cos h - \sin x \sin h - \cos x}{h} \\ &= \lim_{h \to 0} \left[\cos x \left(\underbrace{\frac{\cos h - 1}{h}}_{\to 0}\right) - \sin x \left(\underbrace{\frac{\sin h}{h}}_{\to 1}\right)\right] \\ &= -\sin x \end{aligned} \]

Example 4

Find \( y' \) if \( y = 2\sin x - 5\cos x \)

\[ \begin{aligned} y' &= 2\left[\sin x\right]' - 5\left[\cos x\right]' \\ &= 2\cos x - 5(-\sin x) \\ &= 2\cos x + 5\sin x \end{aligned} \]

Example 5

Find the equation of the tangent line for \( f(x) = 3\sin x \) at \( x = \pi \).

Slope: Need \( f'(\pi) \)

\[ \begin{aligned} f'(x) &= 3[\sin x]' = 3\cos x \\ f'(\pi) &= 3\cos \pi = 3(-1) = -3 \end{aligned} \]

Point: \( (\pi, f(\pi)) = (\pi, 3\sin \pi) = (\pi, 0) \)

\[ \begin{gathered} y - 0 = -3(x - \pi) \\[0.5em] y = -3x + 3\pi \end{gathered} \]

\( e^x \):

\[ \begin{aligned} \frac{d}{dx}(e^x) &= \lim_{h \to 0} \frac{e^{x+h} - e^x}{h} = \lim_{h \to 0} \frac{e^x e^h - e^x}{h} \\ &= e^x \lim_{h \to 0} \frac{e^h - 1}{h}= e^x\cdot 1 = e^x \end{aligned} \]

Theorem \( y = e^x \) is the only that satisfies the following Initial Value Problem (IVP)

\[ \begin{cases} \dfrac{dy}{dx} = y & \leftarrow \quad \frac{d}{dx}(e^x) = e^x \\ y(0) = 1 & \leftarrow \quad e^0 = 1 \end{cases} \]

Example 6

What is \( y'(0) \) if \( y = 5e^x + 7 \)?

\[ \begin{aligned} y' &= 5[e^x]' + [7]'\\y' &= 5e^x + 0 = 5e^x \\ y'(0) &= 5 \cdot e^0 = \boxed{5} \end{aligned} \]

Example 7

When does \( \frac{dy}{dx} = 1 \) if \( y = 72e^x \)?

\[ \begin{aligned} \frac{dy}{dx} = 72 \cdot \frac{d}{dx}(e^x) = 72e^x &\overset{\text{set}}{=} 1 \\ 72e^x &= 1 \\ e^x &= \frac{1}{72} \\ \ln(e^x) &= \ln\left(\frac{1}{72}\right) \\ x &= \ln\left(\frac{1}{72}\right) \\ x &= \ln\left(72^{-1}\right) \\ x &= -\ln 72 \end{aligned} \]

Example 8 (HW 6, # 9)

Find the equation of the tangent line \( f(x) = \frac{x^4}{8} - \frac{16}{x^2} \) at \( x = 2 \)

Slope: Need \( f'(2) \)

\[ \begin{aligned} f(x) &= \frac{1}{8}x^4 - 16x^{-2} \\ f'(x) &= \frac{1}{8}\left[x^4\right]' - 16\left[x^{-2}\right]' \\ &= \frac{1}{8} \cdot 4x^3 - 16 \cdot \left(-2x^{-3}\right) \\ f'(x) &= \frac{1}{2}x^3 + \frac{32}{x^3} \end{aligned} \]\[ \begin{aligned} f'(2) &= \frac{1}{2}\cdot 8 + \frac{32}{8} \\ &= 4 + 4 = 8 \end{aligned} \]

y-coord:

\[ f(2) = \frac{2^4}{8} - \frac{16}{2^2} = 2 - 4 = -2 \]

Equation:

\[ \begin{aligned} y - f(2) &= f'(2)(x-2) \\ y - (-2) &= 8(x-2) \\ y + 2 &= 8(x-2) \\ y + 2 &= 8x - 16 \\ y &= 8x - 18 \end{aligned} \]