Lesson 6: Basic Rules of Differentiation; Derivatives of Sine and Cosine; Natural Exponential Function

Basic Rules of Differentiation

Constant Rule: For any constant \(c\),

\[\frac{d}{dx}(c) = 0\]

Power Rule: For any real number \(n\),

\[\frac{d}{dx}(x^n) = nx^{n-1}\]

Constant Multiple Rule: Let \(c\) be a constant and \(f(x)\) differentiable. Then,

\[\frac{d}{dx}(cf(x)) = c\frac{d}{dx}(f(x))\]

Sum/Difference Rule: Suppose \(f(x)\) and \(g(x)\) are differentiable. Then,

\[\frac{d}{dx}(f(x) \pm g(x)) = \frac{d}{dx}(f(x)) \pm \frac{d}{dx}(g(x))\]

Proof of the Constant Rule

Let's use the definition of a derivative from lesson 5. Let \(f(x) = c\). Then,

\[\frac{d}{dx}(c) = f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{c - c}{h} = \lim_{h \to 0} 0 = 0\]

Important: Recall that \(x^{-n} = \frac{1}{x^n}\) and \(x^{n/m} = \sqrt[m]{x^n}\).

Examples

Example 1: Finding Derivatives Using Basic Rules

Find the derivatives of the following functions.

(a) \(f(x) = 75\)

Solution:

Using the constant rule:

\[f'(x) = \frac{d}{dx}(75) = 0\]

(b) \(f(x) = x^3\)

Solution:

Using the power rule:

\[f'(x) = \frac{d}{dx}(x^3) = 3x^{3-1} = 3x^2\]

(c) \(f(x) = \frac{1}{x^4}\)

Solution:

First, rewrite as \(f(x) = x^{-4}\). Then, using the power rule:

\[f'(x) = \frac{d}{dx}(x^{-4}) = -4x^{-4-1} = -4x^{-5} = -\frac{4}{x^5}\]

(d) \(f(x) = \sqrt[3]{x^2}\)

Solution:

First, rewrite as \(f(x) = x^{2/3}\). Then, using the power rule:

\[f'(x) = \frac{d}{dx}(x^{2/3}) = \frac{2}{3}x^{2/3-1} = \frac{2}{3}x^{-1/3} = \frac{2}{3x^{1/3}} = \frac{2}{3\sqrt[3]{x}}\]

(e) \(f(x) = 4x^3\)

Solution:

Using the constant multiple rule and power rule:

\[f'(x) = \frac{d}{dx}(4x^3) = 4\frac{d}{dx}(x^3) = 4\cdot 3x^{3-1} = 4 \cdot 3x^2 = 12x^2\]

(f) \(f(x) = x^5 + 2x - 3\)

Solution:

Using the sum/difference rule, power rule, and constant rule:

\[f'(x) = \frac{d}{dx}(x^5 + 2x - 3) = \frac{d}{dx}(x^5) + \frac{d}{dx}(2x) - \frac{d}{dx}(3)= 5x^4 + 2 \cdot 1 - 0 = 5x^4 + 2\]

Derivatives of Sine and Cosine

Derivatives of Trigonometric Functions: We note that

\[\frac{d}{dx}\sin(x) = \cos(x) \quad \text{and} \quad \frac{d}{dx}\cos(x) = -\sin(x)\]
Example 2: Derivative of a Trigonometric Function

Suppose \(f(x) = 2\sin(x) - 3\cos(x)\). Find \(f'(x)\).

Solution:

Using the sum/difference rule and constant multiple rule:

\[\begin{align}f'(x) &= \frac{d}{dx}(2\sin(x) - 3\cos(x))\\ &= \frac{d}{dx}(2\sin(x)) - \frac{d}{dx}(3\cos(x))\\ &= 2\frac{d}{dx}(\sin(x)) - 3\frac{d}{dx}(\cos(x))\\ &= 2\cos(x) - 3\cdot (-\sin(x))\\ &= 2\cos(x) + 3\sin(x)\end{align}\]
Example 3: Equation of a Tangent Line

Find the equation of the tangent line to \(f(x) = 2\sin(x) - 3\cos(x)\) at \(x = \frac{\pi}{2}\).

Solution:

From example 2, we know that \(f'(x) = 2\cos(x) + 3\sin(x)\).

The slope of the tangent line at \(x = \frac{\pi}{2}\) is:

\[f'\left(\frac{\pi}{2}\right) = 2\cos\left(\frac{\pi}{2}\right) + 3\sin\left(\frac{\pi}{2}\right) = 2\cdot 0 + 3\cdot 1 = 3\]

The point on the tangent line is \(\left(\frac{\pi}{2}, f\left(\frac{\pi}{2}\right)\right)\):

\[f\left(\frac{\pi}{2}\right) = 2\sin\left(\frac{\pi}{2}\right) - 3\cos\left(\frac{\pi}{2}\right) = 2\cdot 1 - 3\cdot 0 = 2\]

So the point is \(\left(\frac{\pi}{2}, 2\right)\).

Using the point-slope form \(y - y_1 = m(x - x_1)\):

\[y - 2 = 3\left(x - \frac{\pi}{2}\right)\] \[y = 3x - \frac{3\pi}{2} + 2\] \[y = 3x + \frac{4 - 3\pi}{2}\]

Derivative of the Natural Exponential Function

Derivative of the Exponential Function: We note that

\[\frac{d}{dx}(e^x) = e^x\]
Example 4: Derivative of an Exponential Function

Suppose \(f(x) = 7e^x\). Find \(f'(x)\).

Solution:

Using the constant multiple rule:

\[f'(x) = \frac{d}{dx}(7e^x) = 7\frac{d}{dx}(e^x) = 7e^x\]

Practice Problems

Practice Problem 1

Suppose \(f(x) = (x-1)(x+3)\). Find all the values of \(x\) so that \(f'(x) = 1\).

Solution:

First, expand the function:

\[f(x) = x^2 + 3x - x - 3 = x^2 + 2x - 3\]

Find the derivative:

\[f'(x) = 2x + 2\]

Set the derivative equal to 1 and solve:

\[2x + 2 = 1\] \[2x = -1\] \[x = -\frac{1}{2}\]
Practice Problem 2

Suppose \(f(x) = \frac{x^2 - 2x^{3/2}}{\sqrt{x}}\). Find \(f'(x)\).

Solution:

First, simplify by dividing each term by \(\sqrt{x} = x^{1/2}\):

\[f(x) = \frac{x^2}{\sqrt{x}} - \frac{2x^{3/2}}{\sqrt{x}} = \frac{x^2}{x^{1/2}} - \frac{2x^{3/2}}{x^{1/2}}= x^{2-1/2} - 2x^{3/2-1/2} = x^{3/2} - 2x\]

Now find the derivative using the power rule:

\[f'(x) = \frac{d}{dx}(x^{3/2} - 2x) = \frac{3}{2}x^{3/2-1} - 2= \frac{3}{2}x^{1/2} - 2 = \frac{3\sqrt{x}}{2} - 2\]