Before introducing new differentiation rules, let's review the basic rules we already know:
| Rule | Formula |
|---|---|
| Constant Rule | \(\frac{d}{dx}(c) = 0\) for any constant \(c\) |
| Power Rule | \(\frac{d}{dx}(x^n) = nx^{n-1}\) for any \(n\) |
| Exponential Rule | \(\frac{d}{dx}(e^x) = e^x\) |
| Constant Multiple Rule | \(\frac{d}{dx}[c \cdot f(x)] = c \cdot \frac{d}{dx}[f(x)]\) |
| Sum Rule | \(\frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]\) |
| Difference Rule | \(\frac{d}{dx}[f(x) - g(x)] = \frac{d}{dx}[f(x)] - \frac{d}{dx}[g(x)]\) |
Find \(\frac{d}{dx}\left[5e^x - \frac{1}{\sqrt{x}} + 11x^5 + 3x^{5/8} - 124\right]\)
Solution:
Apply the rules term by term:
\[= 5\frac{d}{dx}(e^x) - \frac{d}{dx}(x^{-1/2}) + 11\frac{d}{dx}(x^5) + 3\frac{d}{dx}(x^{5/8}) - \frac{d}{dx}(124)\] \[= 5e^x + \frac{1}{2}x^{-3/2} + 55x^4 + \frac{15}{8}x^{-3/8} - 0\] \[= 5e^x + \frac{1}{2x^{3/2}} + 55x^4 + \frac{15x^{-3/8}}{8}\]Previously, we learned how to differentiate sums and differences of functions. But what about products? The derivative of a product is NOT simply the product of the derivatives.
Verification that \(\frac{d}{dx}[f(x)g(x)] \neq f'(x)g'(x)\):
Consider \(f(x) = x\) and \(g(x) = x\), so \(f'(x) = 1\) and \(g'(x) = 1\).
We know \(\frac{d}{dx}[x \cdot x] = \frac{d}{dx}[x^2] = 2x\)
But \(f'(x) \cdot g'(x) = 1 \cdot 1 = 1 \neq 2x\)
Product Rule:
If \(f(x)\) and \(g(x)\) are differentiable functions, then:
\[\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)\]In words: the derivative of a product equals the derivative of the first times the second plus the first times the derivative of the second.
Let \(f(x) = x\) and \(g(x) = x\), so \(f'(x) = 1\) and \(g'(x) = 1\).
Let \(f(x) = x^2\) and \(g(x) = x\), so \(f'(x) = 2x\) and \(g'(x) = 1\).
Find \(\frac{d}{dx}[(7x + x^3)(2x - 3x^2)]\)
Solution:
Let \(f(x) = 7x + x^3\) and \(g(x) = 2x - 3x^2\)
Then \(f'(x) = 7 + 3x^2\) and \(g'(x) = 2 - 6x\)
Using the Product Rule:
\[\frac{d}{dx}[(7x + x^3)(2x - 3x^2)] = (7 + 3x^2)(2x - 3x^2) + (7x + x^3)(2 - 6x)\] \[= 14x - 21x^2 + 6x^3 - 9x^4 + 14x - 42x^2 + 2x^3 - 6x^4\] \[= 28x - 63x^2 + 8x^3 - 15x^4\]Find \(\frac{d}{dx}[e^x(x^2 + 3x - 7)]\)
Solution:
Let \(f(x) = e^x\) and \(g(x) = x^2 + 3x - 7\)
Then \(f'(x) = e^x\) and \(g'(x) = 2x + 3\)
Using the Product Rule:
\[\frac{d}{dx}[e^x(x^2 + 3x - 7)] = e^x(x^2 + 3x - 7) + e^x(2x + 3)\] \[= e^x(x^2 + 3x - 7 + 2x + 3)\] \[= e^x(x^2 + 5x - 4)\]Just as the derivative of a product is not the product of derivatives, the derivative of a quotient is not the quotient of derivatives.
Verification that \(\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] \neq \frac{f'(x)}{g'(x)}\):
Consider \(\frac{d}{dx}\left[\frac{1}{x}\right] = \frac{d}{dx}[x^{-1}] = -x^{-2} = -\frac{1}{x^2}\)
But if we (incorrectly) divided derivatives: \(\frac{\frac{d}{dx}(1)}{\frac{d}{dx}(x)} = \frac{0}{1} = 0 \neq -\frac{1}{x^2}\)
Quotient Rule:
If \(f(x)\) and \(g(x)\) are differentiable functions and \(g(x) \neq 0\), then:
\[\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}\]In words: the derivative of a quotient equals (derivative of numerator times denominator minus numerator times derivative of denominator) all divided by denominator squared.
Find \(\frac{d}{dx}\left[\frac{x^3}{x}\right] = \frac{d}{dx}[x^2] = 2x\)
Solution using Quotient Rule:
Let \(N = x^3\) (numerator) and \(D = x\) (denominator)
Then \(N' = 3x^2\) and \(D' = 1\)
\[\frac{d}{dx}\left[\frac{x^3}{x}\right] = \frac{3x^2 \cdot x - x^3 \cdot 1}{x^2} = \frac{3x^3 - x^3}{x^2} = \frac{2x^3}{x^2} = 2x \checkmark\]Alternative using Product Rule:
Rewrite as \(\frac{d}{dx}[x^3 \cdot \frac{1}{x}] = \frac{d}{dx}[x^3 \cdot x^{-1}]\)
\[= 3x^2 \cdot \frac{1}{x} + x^3 \cdot (-\frac{1}{x^2}) = \frac{3x^2}{x} - \frac{x^3}{x^2} = 3x - x = 2x \checkmark\]Find \(\frac{d}{dx}\left[\frac{7x - x^2}{3x + 2x^2}\right]\)
Solution:
Let \(N = 7x - x^2\) and \(D = 3x + 2x^2\)
Then \(N' = 7 - 2x\) and \(D' = 3 + 4x\)
\[\frac{d}{dx}\left[\frac{7x - x^2}{3x + 2x^2}\right] = \frac{(7 - 2x)(3x + 2x^2) - (7x - x^2)(3 + 4x)}{(3x + 2x^2)^2}\] \[= \frac{21x + 14x^2 - 6x^2 - 4x^3 - 21x - 28x^2 + 3x^2 + 4x^3}{(3x + 2x^2)^2}\] \[= \frac{-11x^2}{(3x + 2x^2)^2}\]Find \(\frac{d}{dx}\left[\frac{5xe^x}{3x - 1}\right]\)
Solution:
For the numerator \(N = 5xe^x\), we need the Product Rule:
\[N' = \frac{d}{dx}(5x) \cdot e^x + 5x \cdot \frac{d}{dx}(e^x) = 5e^x + 5xe^x\]For the denominator \(D = 3x - 1\), we have \(D' = 3\)
Using the Quotient Rule:
\[\frac{d}{dx}\left[\frac{5xe^x}{3x - 1}\right] = \frac{(5e^x + 5xe^x)(3x - 1) - (5xe^x)(3)}{(3x - 1)^2}\] \[= \frac{15xe^x + 15x^2e^x - 5e^x - 5xe^x - 15xe^x}{(3x - 1)^2}\] \[= \frac{e^x(15x^2 - 5x - 5)}{(3x - 1)^2}\]Find \(\frac{d}{dx}\left[\frac{7x^2e^x - 8x^4}{e^x + 5}\right]\)
Solution:
For the numerator \(N = 7x^2e^x - 8x^4\), using the Product Rule on the first term:
\[N' = \frac{d}{dx}(7x^2) \cdot e^x + 7x^2 \cdot \frac{d}{dx}(e^x) - \frac{d}{dx}(8x^4)\] \[= 14xe^x + 7x^2e^x - 32x^3\]For the denominator \(D = e^x + 5\), we have \(D' = e^x\)
Using the Quotient Rule:
\[\frac{d}{dx}\left[\frac{7x^2e^x - 8x^4}{e^x + 5}\right] = \frac{(14xe^x + 7x^2e^x - 32x^3)(e^x + 5) - (7x^2e^x - 8x^4)(e^x)}{(e^x + 5)^2}\]Connection Between Quotient Rule and Product Rule:
The Quotient Rule can be thought of as an application of the Product Rule:
\[\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{d}{dx}\left[f(x) \cdot \frac{1}{g(x)}\right]\]Using the Product Rule on \(f(x) \cdot [g(x)]^{-1}\):
\[= f'(x) \cdot \frac{1}{g(x)} + f(x) \cdot \frac{d}{dx}\left[\frac{1}{g(x)}\right]\] \[= \frac{f'(x)}{g(x)} - \frac{f(x)g'(x)}{[g(x)]^2} = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}\]We can take the derivative of a function multiple times. Each successive derivative is called a higher order derivative.
Let \(f(x) = 7e^x + x^3 + 3x^6\)
First Derivative:
\[\frac{df}{dx} = f'(x) = \frac{d}{dx}[7e^x + x^3 + 3x^6] = 7e^x + 3x^2 + 18x^5\]Second Derivative (derivative of the first derivative):
Notation: \(\frac{d^2f}{dx^2}\) or \(f''(x)\)
\[\frac{d^2f}{dx^2} = \frac{d}{dx}[f'(x)] = \frac{d}{dx}[7e^x + 3x^2 + 18x^5] = 7e^x + 6x + 90x^4\]Third Derivative (derivative of the second derivative, or second derivative of the first derivative):
Notation: \(\frac{d^3f}{dx^3}\) or \(f'''(x)\)
\[\frac{d^3f}{dx^3} = \frac{d}{dx}[f''(x)] = \frac{d}{dx}[7e^x + 6x + 90x^4] = 7e^x + 6 + 360x^3\]nth Derivative Notation:
The \(n\)th derivative (taking the derivative \(n\) times) can be written as:
Examples: \(\frac{d^{17}f}{dx^{17}}\) or \(f^{(17)}(x)\) represents the 17th derivative
Warning: Notation Ambiguity
Be careful with notation! The superscript means different things in different contexts:
These are completely different! \(f^{(n)} \neq [f(x)]^n\)