MAC 2311C - Exam 2

Derivative of ln(f(x))

f'(x)/f(x)

Product Rule

d/dx [f(x)*g(x)] = f(x)g'(x) + f'(x)g(x)

Quotient Rule

d/dx [f(x)/g(x)] = [g(x)f'(x) - g'(x)f(x)]/[g(x)²]

Chain Rule

d/dx [f(g(x)] = f'(g(x))*g'(x)

cos(pi)

-1

sin(pi)

0

Equation of a tangent line

1.) Find f'(x)

2.) find dy/dx

3.) plug given point into f'(x) (gives slope)

4.) plug slope and coordinates into y-y1=m(x-x1)

Implicit Differentiation

Differentiating both sides of the equation with respect to x and then solving the resulting equation for dy/dx

Second Derivative

1.) Find first derivative with respect to x

2.) Find second derivative with respect to dy/dx... use quotient/chain rule if needed.

3.) Simplify/Combine like terms

(b^x)'

(b^x)(ln b)

Linear Approximation

Y=f(a)+f'(a)(x-a), where f'(a) is the slope of the tangent.

Linearization

L(x)=f(a) + f'(a)(x-a)

Differential

dy=f'(a)d(x) or f(x)=f(a)+dy

Use Linear Approximation to Estimate:

1.) Identify f(x), x, and a value close to x, "a."

2.) Find f'(x)

3.) To approximate x, plug "a" value into linear approximation equation, Y=f(a)+f'(a)(x-a)

4.) Write the linearization equation. Simplify.

5.) Find the differential by substituting proper values into dy=f'(a)d(x).

Identifying Linearization of a Composite Function

1.) write out the linearization equation for a composite function, which is

l(x)= f(g(x))+f(g(x))'(x-a)

2.) Understand that for composite functions, f'(a) becomes f(g(x))', which will need to be simplified using chain rule: f'(g(x))*g'(x).

3.) Use value given in problem and the table to find f(g(x))'. Continue to find l(x) normally.

Absolute Maximum

Highest Elevation, or highest endpoint, if applicable.

Absolute Minimum

Lowest Elevation, or lowest endpoint, if applicable.

Absolute Extremum

absolute min and max

Local Max

Max @ point in the given interval

Local Min

Min @ point in the given interval

Guidelines for Finding Absolute Extrema on a Closed Interval

1.) Find the critical values of f in (a,b).

2.) Evaluate f at each critical numver in (a,b)

3.) Evaluate f at each endpoint of [a,b]

4.) the least of these values is the minimum or absolute maxiumum. The greatest is the maximum or absolute maximum.

Extreme Value Theorem

If f is continuous on the closed interval [a,b] then there are both an absolute minimum and absolute maximum on [a,b].

Function with both an absolute max and absolute min on R.

F(x)=sin x

Function with an absolute max but not an absolute min on R.

F(x)=1/1+x^2

Function with absolute min but not an absolute max on R.

F(x)=2^x

Function with neither an absolute max or absolute min.

F(x)=arctan x

Fermat's Theorem

If f has a local extremum @ c, and f is differentiable @ c, then f'(c)=0.

Critical Point

Let c be an interior point on the domain of f. The critical point(s) of f are f'(c)=0 or f'(c)=DNE/undefined

Find Absolute Max & Min on Closed Interval

1.) Find first derivative of given equation.

2.) find when f'(x)=0 and f'(x)=DNE/undefined

3.) plug in a and b from [a,b] into original equation.

4.) plug critical points into original equation.

Let f(x) be a differentiable function on a closed interval with x=a being one of the endpoints of the interval. If f'(a)>0 then

F could have an absolute max or absolute min at x=a

Rollers Theorem

If f is continuous on [a,b], differentiable on (a,b), and f(a)=f(b), then f'(c)=0 for some point c on (a,b).

Consider the function f(x)=1-x^2/3. We find that f(-1)=f(1)=0, but there is no c in the interval (-1,1) such that f'(c)=0.

This does not contradict Rolle's Theorem because f'(0) does not exist, and so f is not differentiable on (-1,1).

Mean Value Theorem

If f: [a,b] → R is continuous on [a,b] and differentiable on (a,b), then there exists a point c ∈ (a,b) where f'(c) = [f(b) -f(a)]/[b-a]. Alternative formula: f(b)-f(a)=f'(c)(b-a) for some point c in (a,b).

First Corollary of the Mean Value Theorem

If f is differentiable over an interval I and f'(x)=0 for all x∈ I, then f is a constant function on I.

Second Corollary of the Mean Value Theorem

If f and g are differentiable over an interval I and f'(x)=g'(x) for all x∈ I, then there is a constant C such that f(x)=g(x)+C

Third Corollary of the Mean Value Theorem

Let f be continuous over the closed interval [a,b], and differentiable over the open interval (a,b):

(I) If f(x)>0 for all x∈ (a,b) then f is an increasing function over [a,b].

(II) if f'(x)<0 for all x∈ (a,b) then f is a decreasing function over [a,b].

Derivatives Affecting Graph Shape

If f'(x)>0, f is increasing on interval I.

If f'(x),0, f is decreasing on interval I.

If f''(x)>0 on I, then f is concave up on interval I. :)

If f''(x)<0 on I, then f is concave down on interval I. :(

Inflection Point

Let f be a function differentiable in a neighborhood of a point c. If the sign of f" changes c, then c is called inflection point.