notes/education/math/MATH1210 (calc 1)/Derivatives.md

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.

Point Slope Formula (Review)

y - y_1 = m(x-x_1)

Given that m = f'(a) and that (x_1, y_1) = (a, f(a)), you get the equation:

y - f(a) = f'(a)(x - a)

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 f with 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

Constant Rule

The derivative of a constant is always zero.

\dfrac{d}{dx}[c] = 0

For example, the derivative of the equation f(x) = 3 is 0.

Derivative of x

The derivative of x is one.

For example, the derivative of the equation f(x) = x is 1, and the derivative of the equation f(x) = 3x is 3.

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 = a^x*(\ln(a))

for all a > 0

Chain Rule

\dfrac{d}{dx} f(g(x)) = f'(g(x))*g'(x)

Examples

Given the function (x^2+3)^4, find the derivative.

Using the chain rule, the above function might be described as f(g(x)), where f(x) = x^4, and g(x) = x^2 + 3).

1. First find the derivative of the outside function function (f(x) = x^4):

\dfrac{d}{dx} (x^2 +3)^4 = 4(g(x))^3 ...

2. Multiply that by the derivative of the inside function, g(x), or x^2 + 3.

\dfrac{d}{dx} (x^2 + 3)^4 = 4(x^2 + 3)^3 * (2x)

Apply the chain rule to x^4

If we treat the above as a function along the lines of f(x) = (x)^4, and g(x) = x, then the chain rule can be used like so:

4(x)^3 * x

Trig Functions

\lim_{x \to 0} \dfrac{\sin x}{x} = 1 \lim_{x \to 0} \dfrac{\cos x - 1}{x} = 0

Sine

f'(x) = \lim_{h \to 0} \dfrac{\sin(x + h) - sin(x)}{h}

Using the sum trig identity, \sin(x + h) can be rewritten as \sin x \cos h + \cos x \sin h.

This allows us to simplify, ultimately leading to:

\dfrac{d}{dx} \sin x = \cos x

Cosine

\dfrac{d}{dx} \cos x = -\sin x

Tangent

\dfrac{d}{dx} \tan x = \sec^2x

Secant

\dfrac{d}{dx} \sec x = \sec x * \tan x

Cosecant

\dfrac{d}{dx} \csc x = -\csc x \cot x

Cotangent

\dfrac{d}{dx} \cot x = -\csc^2 x

Implicit Differentiation

• There's a reason differentials are written like a fraction
• \dfrac{d}{dx} x^2 = \dfrac{d(x^2)}{dx}, or, "the derivative of x^2 with respect to $x$"
• \dfrac{d}{dx} x = \dfrac{dx}{dx} = 1 : The derivative of x with respect to x is one
• \dfrac{d}{dx} y = \dfrac{dy}{dx} = y'
• Given the equation y = x^2, \dfrac{d}{dx} y = \dfrac{dy}{dx} = 2x.

Examples

Differentiate f(x) = 4\sqrt[3]{x} - \dfrac{1}{x^6}

2. f(x) = 4\sqrt[3]{x} = \dfrac{1}{x^6}
3. = 4x^\frac{1}{3} - x^{-6}
4. f'(x) = \dfrac{1}{3} * 4x^{-\frac{2}{3}} -(-6)(x^{-6-1})
5. = 4x^{-2-\frac{2}{3}} + 6x^{-7}
6. = \dfrac{4}{3\sqrt[3]{x^2}} + \dfrac{6}{x^7}