Calculus Unit 3
Theorem: Differentiability Implies Continuity
Let f(x) be a function and a be in its domain. If f(x) is differentiable at a, then f is continuous at a.
Point slope equation of a line
y-y1=m(x-x1)
Absolute Value Limit
|a| = {a if a ≥ 0, −a if a < 0.
The Quotient Rule in words
(numerator/denominator)′ = ((numerator′ * denominator) - denominator′ * numerator)/(denominator)^2 "lo d hi minus hi d lo over the square of what's beLO"
Finding the Derivative tips
-Generally, the algebra isn't that bad -Look for when you can keep things factored (preventing unnecessary distribution) -When you have radicals, you can also rationalize numerators. -if the limit doesn't exist then the derivative doesn't exist either.
The position of an object at any time t (in hours) is given by the function s(t). Determine when the object is moving to the right and when the object is moving to the left.
1) Get derivative to get velocity velocity is positive = the object is moving off to the right velocity is negative = the object is moving to the left -Make sure to factor the derivative for the next step 2) Find where t = 0. IVT states that if the polynomial ever changes sign then it must have first gone through zero. So, if we knew where the derivative was zero we would know the only points where the derivative might change sign. 3) Using the new points, divide the number line into distinct regions In each of these regions we know that the derivative will be the same sign. Recall the derivative can only change sign at the two points that are used to divide the number line up into the regions. Therefore, all that we need to do is to check the derivative at a test point in each region and the derivative in that region will have the same sign as the test point. intervals where derivative is positive = moving right intervals where derivative is negative = moving left
Find the equation of the line tangent to the graph
1. First find the slope of the tangent line. 2. Next, find a point on the tangent line (try substituting the number used to solve for the tangent line's slope). 3. Using the point-slope equation of the line with the slope and the point to obtain the line.
Summary on Differentiability
1. We observe that if a function is not continuous, it cannot be differentiable, since every differentiable function must be continuous. However, if a function is continuous, it may still fail to be differentiable. 2. We saw that f(x)=|x| failed to be differentiable at 0 because the limit of the slopes of the tangent lines on the left and right were not the same. Visually, this resulted in a sharp corner on the graph of the function at 0. From this we conclude that in order to be differentiable at a point, a function must be "smooth" at that point. 3. As we saw in the example of f(x)=x√3, a function fails to be differentiable at a point where there is a vertical tangent line. 4. As we saw with f(x)= {x sin(1/x) if x≠0 {0 if x=0 a function may fail to be differentiable at a point in more complicated ways as well.
The Constant Rule
Let c be a constant. If f(x) = c , then f′(c) = 0. Alternatively, we may express this rule as d/dx (c) = 0
The Quotient Rule
Let f(x) and g(x) be differentiable functions. Then d/dx(f(x)/g(x)) = (d/dx(f(x))⋅g(x)−d/dx(g(x))⋅f(x))/(g(x))^2. That is, if j(x) = f(x)/g(x), then j′(x) = (f′(x)g(x)−g′(x)f(x))/(g(x))^2. SUPER IMPORTANT RULE FOR CALCULATING ADVANCED DERIVATIVES
The Product Rule
Let f(x) and g(x) be differentiable functions. Then d/dx(f(x)g(x)) = d/dx(f(x))⋅g(x) + ddx(g(x))⋅f(x). That is, if j(x) = f(x)g(x), then j′(x) = f′(x)g(x)+g′(x)f(x). This means that the derivative of a product of two functions is the derivative of the first function times the second function plus the derivative of the second function times the first function.
Tangent line
Let f(x) be a function defined in an open interval containing a. The tangent line to f(x) at a is the line passing through the point (a,f(a)) having slope mtan=limx→a (f(x) − f(a))/(x − a) provided this limit exists. Equivalently, we may define the tangent line to f(x) at a to be the line passing through the point (a,f(a)) having slope mtan=limh→0 (f(a+h) − f(a))/h provided this limit exists.
derivative
Let f(x) be a function defined in an open interval containing a. The derivative of the function f(x) at a, denoted by f′(x), is defined by f′(x)=limx→a f(x)−f(a)/x−a provided this limit exists. Alternatively, we may also define the derivative of f(x) at a as f′(a)=limh→0 f(x+h)−f(x)/h. Note that here we replaced all the a's with x's to acknowledge the fact that the derivative is really a function as well. We often "read" f′(x) as "f prime of x". Many applications across many disciplines (ex. slope of the line tangent to a function at a point, velocity and acceleration in physics, marginal profit functions in business, growth rates in biology).
Secant lines approaching
Pictured: In Figure (a) we see that, as the values of x approach a, the slopes of the secant lines provide better estimates of the rate of change of the function at a. Furthermore, the secant lines themselves approach the tangent line to the function at a, which represents the limit of the secant lines. Similarly, Figure (b) shows that as the values of h get closer to 0, the secant lines also approach the tangent line. The slope of the tangent line at a is the rate of change of the function at a, as shown in Figure (c). The secant lines approach the tangent line (shown in green) as the second point approaches the first.
Difference quotient
Slope of a secant line f(a+h)-f(a)/h or f(x)-f(a)/x-a a must be a value within the interval I. h ≠ 0, x≠a.
Derivative Function
The derivative function gives the derivative of a function at each point in the domain of the original function for which the derivative is defined. Let f be a function. The derivative function, denoted by f′, is the function whose domain consists of those values of x such that the following limit exists: f′(x)=limh→0 f(x+h)−f(x)/h. A function f(x) is said to be differentiable at a if f(a) exists. More generally, a function is said to be differentiable on S if it is differentiable at every point in an open set S, and a differentiable function is one in which f′(x) exists on its domain.
Constant Multiple Rule
The derivative of a constant c multiplied by a function f is the same as the constant multiplied by the derivative: d/dx(kf(x)) = kd/dx(f(x)) that is, for j(x) = kf(x), j′(x) = kf′(x).
Difference Rule
The derivative of the difference of a function f and a function g is the same as the difference of the derivative of f and the derivative of g: d/dx (f(x)−g(x)) = d/dx(f(x)) − d/dx(g(x)) that is, for j(x) = f(x)−g(x), j′(x) = f′(x)−g′(x).
The Derivatives of sin x and cos x
The derivative of the sine function is the cosine and the derivative of the cosine function is the negative sine. d/dx(sinx)= cos x d/dx(cosx)= −sin x
Sum Rule
The derivative of the sum of a function f and a function g is the same as the sum of the derivative of f and the derivative of g. d/dx (f(x)+g(x)) = d/dx(f(x)) + d/dx(g(x)) that is, for j(x) = f(x)+g(x), j′(x) = f′(x)+g′(x).
Instantaneous rate of change
The instantaneous rate of change of a function f(x) at a point is its derivative evaluated at a or f′(a).
differentiation
The process of finding a derivative
Radical rule
When you see radicals you should always first convert the radical to a fractional exponent and then simplify exponents as much as possible.
The Power Rule
d/dx x^n = nx^(n-1) or f′(x) = nx^(n−1).
Derivative of cot(x)
d/dx(cotx) = −csc^2x
Derivative of cse(x)
d/dx(cscx) = −csc x cot x.
Derivative of sec(x)
d/dx(secx) = sec x tan x
Derivative of the Sine Function with h = 0.01
d/dx(sinx) ≈ (sin(x+0.01)−sinx)/0.01 Very close to the graph of the cosine function...
Derivative of tan(x)
d/dx(tanx) = sec^2x
Power Rule Demonstration
d/dx(x^2) = 2x and d/dx(x^(1/2)) = 1/2x^(−1/2).
Continuity does not imply differentiability
ex. f(x) = |x| is continuous at x=0 but is not differentiable at x=0 as the function's two one-sided limits are not equal.
Why we cannot simply take the quotient of the derivatives
keep in mind that d/dx(x^2) = 2x, not (d/dx(x^3))/(ddx(x)) = 3x^2/1 = 3x^2.
Two important trigonometric limits
limh→0 sin(h)/h = 1 and limh→0 cos(h-1)/h = 0.
We used the slope of a secant line to a function at a point (a,f(a)) to estimate the ____, or the rate at which one variable changes in relation to another variable
rate of change
Sine of the sum of two angles
sin(x+h) = sin x cos h + cos x sin h.
secant line
straight line joining two points on a function
