Derivative Basics
| Monday, 21 February 2022 | |
| 3-minute read | |
| 623 words | |
Derivatives
The derivative of a function \(f'(x)\) describes the instantaneous slope of a function \(f(x)\) at a given \(x\).
For example, at the constant function \(f(x) = b\), the derivative of the function would be \(f'(x) = 0\) as there is no slope in the function. At the linear function \(f(x) = mx + b\), the derivative of the function would be \(f'(x) = m\) since the slope is constantly \(m\).
To avoid confusion, we will sometimes be using both \(x\) and \(x_0\). It is important to know \(f'(x_0)\) is \textbf{not the function of a line}, but rather is equal to the slope of the line \(f(x)\) at any given \(x_0\).
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The orange line describes the rate of change of the line tangent to \(f(x)\) at \(x_0 = 1\)
The orange line describes the rate of change of the line tangent to \(f(x)\) at \(x_0 = 0.1\)
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The derivative notation \(\frac{d}{dx}[f(x)]\) (differentiation with respect to \(x\)) will be used in most other cases instead, but it is still equivalent to prime notation \(f'(x)\) (\(f\) prime of \(x\)).
Power Rule
How do you find the derivative of any polynomial?
\[ \frac{d}{dx}(ax^n) = anx^2{n-1} \]
So given \(f(x) = x^2\), \(\frac{d}{dx}(f(x)) = 2x\)
Using our function \(f(x) = x^2\) from the previous images, we can determine the slope at any point \(x_0\) with \(f'(x_0) = 2x_0\).
Example
Using the function \(g(x) = 3x^4\), plot the instantaneous rate of change on a table.
We can use the power rule to find the derivative.
\[ \frac{d}{dx}(g(x)) = 3(4)^3 = 12x^3 \]
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| \(x\) | \(g(x)\) | slope at \((x, g(x))\) |
|---|---|---|
| 0 | 0 | \(g'(0) = 0\) |
| 1 | 3 | \(g'(1) = 12\) |
| 2 | 16 | \(g'(2) = 96\) |
| 3 | 243 | \(g'(3) = 324\) |
| 4 | 768 | … |
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Difference quotient
The difference quotient determines the average slope over an interval in a function. The difference quotient derives from the slope formula:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{f(x + h) - f(x)}{x + h - x} = \frac{f(x + h) - f(x)}{h} \]
where \(h\) is the length of the interval or change in \(x\).
The difference quotient yields the slope of the secant line passing the two points. As the value \(h\) approaches 0, the secant line starts to match the tangent line.
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This means the difference quotient can be used to find the derivative.
\[ \lim_{h \to 0}\frac{f(x + h) - f(x)}{h} = f'(x) \]
Example
Find the difference quotient of \(p(x) = 2x^2 - 4x\).
\[ \frac{p(x + h) - p(x)}{h} \]
\[ \frac{2(x + h)^2 - 4(x + h) - 2x^2 + 4x}{h} \]
\[ \frac{2x^2 + 4xh + 2h^2 - 4x - 4h - 2x^2 + 4x}{h} \]
\[ \frac{4xh + 2h^2 - 4h}{h} \]
\[ \frac{h(4x + 2h - 4)}{h} \]
\[ 4x + 2h - 4 \]
Derivative from difference quotient
To find the difference quotient, \(h\) must equal \(0\) that \((x, f(x))\) and \((x + h, f(x + h))\) are the same point. It is important they are the same point since a tangent line can only pass a curve at exactly one point.
Using the difference quotient from the previous example, let \(h = 0\).
\[ p'(x) = 2x + 4(0) - 1 = 4x - 4 \]
Using the power rule, we get the same result: \(\frac{d}{dx}p(x) = anx^{n-1} = 2(2)(x)^1 - 4x^0 = 4x - 4\)
Equation of the tangent line
Since the derivative is the slope of the line tangent to \(f(x)\) passing through \((x_0, f(x_0))\), we can use \(f(x_0)\) to determine the instantaneous slope at \(x_0\).
We can then use the point-slope form of a linear equation \(y - y_0 = m(x - x_0)\).
\[ y - f(x_0) = f'(x)(x - x_0) \]