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A derivative can be used to describe the rate of change at a single point, or the instantaneous velocity.
The formula used to calculate the average rate of change looks like this:
\dfrac{f(b) - f(a)}{b - a}
Interpreting it, this can be described as the change in y over the change in x.
- Speed is always positive
- Velocity is directional
As the distance between the two points a and b grow smaller, we get closer and closer to the instantaneous velocity of a point. Limits are suited to describing the behavior of a function as it approaches a point.
If we have the coordinate pair (a, f(a)), and the value h is the distance between a and another x value, the coordinates of that point can be described as ((a + h, f(a + h)). With this info:
- The slope of the secant line can be described as
\dfrac{f(a + h) - f(a)}{a + h - a}, which simplifies to\dfrac{f(a + h) - f(a)}{h}. - The slope of the tangent line or the instantaneous velocity can be found by taking the limit of the above function as the distance (
h) approaches zero:
\lim_{h \to 0}\dfrac{f(a + h) - f(a)}{h}
The above formula can be used to find the derivative. This may also be referred to as the instantaneous velocity, or the instantaneous rate of change.
Line Types
Secant Line
A Secant Line connects two points on a graph.
A Tangent Line represents the rate of change or slope at a single point on the graph.
Notation
Given the equation y = f(x), the following are all notations used to represent the derivative of f at x:
f'(x)\dfrac{d}{dx}f(x)y'\dfrac{dy}{dx}\dfrac{df}{dx}- "Derivative of
fwith respect to $x$"
Functions that are not differentiable at a given point
- Where a function is not defined
- Where a sharp turn takes place
- If the slope of the tangent line is vertical
Higher Order Derivatives
- Take the derivative of a derivative
Exponential Derivative Formula
Using the definition of a derivative to determine the derivative of f(x) = x^n, where n is any natural number.
f'(x) = \lim_{h \to 0} \dfrac{(x + h)^n - x^n}{h}
- Using pascal's triangle, we can approximate
(x + h)^n
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
- Where
n = 0:(x + h)^0 = 1 - Where
n = 1:(x +h)^1 = 1x + 1h - Where
n = 2:(x +h)^2 = x^2 + 2xh + h^2 - Where
n = 3:(x + h)^3 = 1x^3h^0 + 3x^2h^1 + 3x^1h^2 + 1x^0h^3 = 1x^3 + 3x^2h + 3xh^2 + 1h^3
Note that the coefficient follows the associated level of Pascal's Triangle (1 3 3 1), and $x$'s power decrements, while $h$'s power increments. The coefficients of each pair will always add up to n. Eg, 3 + 0, 2 + 1, 1 + 2, and so on. The second term in the polynomial created will have a coefficient of n.
\dfrac{(x + h)^n - x^n}{h} = \lim_{h \to 0} \dfrac{(x^n + nx^{n-1}h + P_{n3}x^{n-2}h^2 + \cdots + h^n)-x^n}{h} P denotes some coefficient found using Pascal's triangle.
x^n cancels out, and then h can be factored out of the binomial series.
This leaves us with:
\lim_{h \to 0} nx^{n-1} + P_{n3} x^{n-2}*0 \cdots v * 0
The zeros leave us with:
f(x) = n, \space f'(x) = nx^{n-1}
Sum and Difference Rules
\dfrac{d}{dx}(f(x) \pm g(x)) = f'(x) \pm g'(x)
Product Rule
\dfrac{d}{dx} (f(x) * g(x)) = \lim_{h \to 0} \dfrac{f(x +h) * g(x + h) - f(x)g(x)}{h}
This is done by adding a value equivalent to zero to the numerator (f(x + h)g(x) - f(x + h)g(x)):
\dfrac{d}{dx} (f(x) * g(x)) = \lim_{h \to 0} \dfrac{f(x +h) * g(x + h) + f(x + h)g(x) - f(x+h)g(x) - f(x)g(x)}{h}
From here you can factor out f(x + h) from the first two terms, and a g(x) from the next two terms.
Then break into two different fractions:
\lim_{h \to 0} \dfrac{f(x + h)}{1} * \dfrac{(g(x + h) - g(x))}{h)} + \dfrac{g(x)}{1} *\dfrac{f(x + h) - f(x)}{h}
From here, you can take the limit of each fraction, therefore showing that to find the derivative of two values multiplied together, you can use the formula:
\dfrac{d}{dx}(f(x) * g(x)) = f(x) * g'(x) + f'(x)*g(x)
Constant Multiple Rule
\dfrac{d}{dx}[c*f(x)] = c * f'(x)
Quotient Rule
\dfrac{d}{dx}(\dfrac{f(x)}{g(x)}) = \dfrac{f'(x)g(x) -f(x)g'(x)}{(g(x))^2}
Exponential Rule
\dfrac{d}{dx} e^x = e^x
$$ \dfrac{d}{dx}a^x