- StudyBlue
- Florida
- Valencia Community College
- Calculus
- Calculus 2311
- Boustique
- Test 2 Review

Victoria B.

A Constant

f'(x)=0

Advertisement

Product Rule

d/dx uv = uv' + u'v

Quotient Rule

y=f(x)/g(x)

y'=g(x)f'(x)-f(x)g'(x)//g(x)2

"Low d high - high d low, all over low^{2}"

f'(x)=sin(x)

cos(x)

f'(x)=cos(x)

-sin(x)

f'(x)=tan(x)

sec^{2}x = 1/(cos^{2}x)

f'(x)=cot(x)

-csc^{2}(x)

f'(x)=sec(x)

tan(x)sec(x)

f'(x)=csc(x)

-csc(x)cot(x)

Chain Rule

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

Advertisement

Implicit Differentiation

Y is not isolated. Y is treated as a function.

1. find deriv of both sides2. combine dy/dx

3. factor dy/dx

4. solve for dy/dx

y'=dy/dx

Implicit differentiation

Speed

derivative of displacement(S(t))

Acceleration

derivative of speed

Linear approximation

1. Find a function f(x)

2. Find f'(x)

3. Find f(#) to get point and f'(#) to get slope at the point and make equation for tangent line

4. plug in number you're trying to approximate

5. plug in # and slope into y-y_{1}=m(x-x_{1})

The differential

df=f'(a)dx

how to find critical points?

when f'=0 or DNE

Local Maximum

If f'(x) changes from positive to negative at a critical point.

Local Minimum

If f'(x) changes from negative to positive at a critical point.

Absolute maximum

The highest maximum on a graph

absolute minimum

The lowest point over the entire domain of a function or relation.

f(x) has a critical point at x=C if one of the following cases happens

1.) f'(C)=0

2.) f'(C) DNE while f(C) exists

Fermat's Theorem

If f has a local max/min & f'(C) exists, then f'(C)=0. C is a critical # of f.

Extreme Value Theorem

If *f* is continuous on a closed interval [a,b], then *f* attains an absolute maximum value *f(c) *and an absolute minimum value *f(d) *at some numbers *c *and* **d* in [a,b],

To find the absolute max/min:

1. find all critical points within [a,b]

2.evaluate the function at each of the critical points(found in step 1)

3. evaluate the function at the end pts (i.e. (f(a),f(b))

4.-The highest value obtained in step 2 & 3 is absolute max.

-The lowest value obtained in step 2 & 3 is the absolute min

Rolle's Theorem

1. f is continuous on closed interval [a,b]

2. f is differentiable on the open interval (a,b)

3. f(a)=f(b)

Then there is a number c in (a,b) such that f'(c)=0

The Mean Value Theorem

If f(x) is

(1) continuous on [a,b]

(2) differentiable on (a,b,

then there exists a # c in (a,b) such that

f'(c)=__f(b) - f(a)__

b - a

(1) continuous on [a,b]

(2) differentiable on (a,b,

then there exists a # c in (a,b) such that

f'(c)=

b - a

Test for Increasing/Decreasing Functions

Let f be continuous on [a, b] and differentiable on (a, b)

1. If f'(x) > 0 for all x in (a, b), then f is increasing on [a, b]

2. If f'(x) < 0 for all x in (a, b), then f is decreasing on [a, b]

3. If f'(x) = 0 for all x in (a, b), then f is constant on [a, b]

First Derivative Test

Suppose that C is a crit. # of a continuous function f

a) If f' changes from + to - at C, then f has a local max @ C

b) " " from - to + C, then f has a local min @ C

c) If f' does not change sign @ C, then f has no local max or min @ C

Concavity test

a) f"(x)>0 for all x in I, then the graph of f is concave upward on I

b) If f"<0 for all x in I, then the graph of f is concave downward on I

Inflection Point

A Point P on a curve y=f(x) if f is continuous there & the curve changes from concave up to concave down or from down to up at P.

Second Derivative Test

Suppose f'' is continuous near c.

(a) If f'(c) = 0 and f''(c)>0, then f has a local minimum at c.

(b) If f'(c) = 0 and f''(c)<0, then f has a local maximum at c.

(a) If f'(c) = 0 and f''(c)>0, then f has a local minimum at c.

(b) If f'(c) = 0 and f''(c)<0, then f has a local maximum at c.

"The semester I found StudyBlue, I went from a 2.8 to a 3.8, and graduated with honors!"

Jennifer Colorado School of Mines© 2014 StudyBlue Inc. All rights reserved.