af Daniel Mandujano 1 år siden
120
Mere som dette
f'(c)=f(b)-f(a)/b-a
f(b)-f(a)=f'(c)(b-a)
2. f is differentiable on [a,b]
1. f is continuous on a closed interval [a.b]
if f has a local max or min at c, then c is called a critical point of f.
max and min are often referred as EXTREMA
Let c be a number in the domain D of a function f(c) is a local maximum value of f on D if f(c) is greater than or equal to f(x) for all values of x neighborhood around c.
Make sure that you know what you are graphing
Derivative has to exist for it to change
IP is when the signs changes(negative and positive)
Where the first derivative changes signs
Create number line to know where it is positive or negative
Critical points, y'=0
First Derivative
b. Vertical
denominator is 0
a. Horizontal
use limits to find
y-intercepts set x=0
x-intercepts set y=0
log(x) is greater than 0
Even radicals is greater than or equal to 0
Denominator can not equal 0
Test for Concavity
If c is a point at which f''(x) changes sign at c, then f has an inflection point at c.
If f''(x) less than 0 on I, then f is concave down
If f''(x) greater than 0 on I, then f is concave up
if f is ontinuous at c and f changes concavity at c (from up to down or from down to up), then f has an inflection point at c.
if the graph of f lies below all of its tangents on an interval I, then it is said to be concave down.
Definition: If the graph of f lies above all of its tangents on interval I, then it is said to be concave up
If f''(c)=0, then the second derivative test is inconclusive
If f''(c) is less than 0, then f has a local maximum.
Suppose f'' is continuous on an open interval containing c with f'(c)=0
Supposed f is continuous on an interval that contains a critical point c and is differentiable on the interval, except possibly c.
If f' positive on both sides of c or negative on both sides of c, then f has no extreme value at c.
if f' changes from negative to positive as x increases through c (from left to right), then f has a local minimum at c.
If f' changes sign from positive to negative as x increases through c (from left to right), then f has a local maximum at c.
Positive Slope
Physical interpretation, if the signed distance s from the orgin at time t of an object in linear motion is given by the position s=f(t), the derivative f'(t0) is the velocity of the object at time t0
Geometric interpretation if y=f(x), the derivative f'(c) is the slope of the tangent line to the graph of f at point (c,f(c))
A rancher has 400 ft of fence for constructing three adjacent rectangular corrals. One side will be formed by the barn and requires no fence. What dimensions of the corrals will maximize the enclosed area?
400=8x
400/8=x
x=50
4(50)+y=400
y=200
Step 5: Use the First Derivative Test to find the absolute max or min. If necessary, check end points.
A'=400-8x
Step 4: Write the optimization equation expressed with a single variable
A=400x-4x^2
Step 3: Use the constraints to eliminate all but one independent variable.
A=(400-4x)x
Step 2: Identify the Optimization equation. Write it in terms of the variables from the constraints.
Step 1:Read the problem, draw a picture of diagram to repents all quantities and identify the constraints
4x+y=400
dy=f'(x)dx
If y=f(x), where f is differentiable function, then the differential dx is an independent variable . That is, dx can be given the value of any number. The differential dy is then defined in terms of dx by equation.
Implicit Differentiation
Find dy/dx of x^2+y^2=1
x^2(d/dx)+y^2(d/dx)=1(d/dx)
2x+2y=0
2x+2y(dy/dx)=0
2y(dy/dx)=-2x
dy/dx=-2x/2y
Answer: dy/dx=-x/y
this is why the dx/dy notation is important?
The derivative of y with respect to x.
Equations that are not written in terms of one variable are called implicit equations. Some implicit equations cannot be written explicitly.
x^2+y^2=1
y^2+x^2+1
y=+/- square root (x^2+1)
half a circle
f(x)=f(a)+f'(a)(x-a)
f'(x)=f(x)-f(a)/x-a
is called the linear approximation or tangent line approximation of f at a. The linear function whose graph is this tangent line, that is L(x)=f(a)+f'(a)(x-a) is called the linearization of f at a.
a function will fail to be differentiable at a point a if
2. The function is discontinuous at a.
1. There is a casp or kink at a (bounce)
Higher Order Derivatives
f'(x) f''(x) f'''(x) f^4(x)
y' y'' y''' y^4
Dx D^2x D^3x D^nx
dy/dx d^2y/dx^2 d^3y/dx^3 d^ny/dx^n
The derivative of f is f(x) prime=dy/dx=limf(a+h)+f(a)/h
Provided the limit exists and a is in the domain of f. If f(x) prime exists, we say f is differentiable at a. If f is differentiable on every point of an interval I, we say that f is differentiable on I
mtan=limf(a+h)+f(a)/h
h→0
Find the tangent line to the function f(x)=-3x^2-5x+1 at (1,-7), Then find the normal line
First thing you are looking for is f(a+h)
-3(a+h)^2-5(a+h)+1
Alternate definition
f on the interval [a,a+h]
From this we can get the tangent line passing through point (a,f(a))
Using the point-slope formula
mtan=point
mtan=f(a) prime
x→a
d/dx(sechx)=-sech x tanh x
d/dx(coth x)= -csch^2 x
d/dx(csch x)= -cschxcoth x
d/dx(sinhx)=cosh x
d/dx(cosh x)=sinh x
d/dx(tanhx)= sech^2 x
cosh^2x-sinh^2x=1
cosh(x+y)=coshxcoshy+sinhxsinhy
1-tanh^2x=sech^2x
sin^2x+cos^2x=1
cosh(-x)=-cosh(x)
sinh(x+y)=sinhxcoshy+coshysinhx
sinh(-x)=-sinh x
Functions
sinh x = e^x-e^-x/2
Range: (-∞, ∞)
Domain:(-∞, ∞)
tanh x = sinh x/cosh x=e^x-e^-x/e^x+e^-x
Range: (-1,1)
Domain: (-∞,∞)
sech x = 1/cosh x = 2/e^x+e^-x
Range: (0,1]
Domain: (-∞, ∞)
coth x=cosh x/sinh x=e^x+e^-x/e^x-e^-x
Range: (-∞, -1)u(1,∞)
csch x = 1/sinh x = 2/e^x-e^-x
Range: (-∞, 0)u(0,∞)
Domain: (-∞, 0)u(0,∞)
cosh x = e^x+e^-x/2
Range: [1,∞)
Domain: (-∞, ∞)
Example:
y=1/log6x^2
rewrite as 1/lnx^2/ln6
Do KCF(Keep Change and Flip)
ln6/2ln x
1/2(ln6)(ln x)^-1
-ln6/2x(ln x)^2
Using b^ylnb and implicit differentiation of y=logbX
we can find
logbx=ln x/lnb
d/dx(log b x=1/xlnb
d/dx(cot^-1 x) = -1/1+x^2
d/dx(sec^-1 x) = 1/x√1-x^2
d/dx(sin^-1 x) = 1/√1-x^2
d/dx(cos^-1 x) = -1/√1-x^2
d/dx(csc^-1 x) = -1/x√1-x^2
d/dx(tan^-1 x) = 1/1+x^2
You should distinguish carefully between the Power Rule, where the base is variable and the exponent is constant, and the rule for differentiating exponential functions, where the base is constant and the exponent is variable.
b^x=b^xln b
Power Rule
x^n=nx^n-1
Steps
3. Solve the resulting equation for y' and replace y by f(x)
2. Differentiate implicitly with respect to x
!. Take natural logarithms of both sides of an equation y= f(x) and use the Laws of Logarithms to expand the expression.
The calculation of derivatives of complicated functions involving products, or powers can be simplified by taking logarithms
If we don't do logarithmic differentiation, we have to use Quotient Rule or Product Rule.
Find the derivative of y=x^3(3^x)
3x^2(3^x)+(ln3)3^x(x^3)
x^2(3^x)(3+xln3)
s= 3^x
s'=(ln3)3^x
f=x^3
f'=3x^2
e^xlnb
ln b e ^xln b
d/dx(b^x)
ln b(b^x)
b cannot equal 1
b is greater than 0
Rules of exponents can redefine b^x=e^ln bx=e^xlnb
use this variation to differentiate b^x
Use definition and implicit differentiation to find d/dx(ln x)
y=ln x
(dy/dx)e^y=x(dy/dx)
dydx(y^y)=x
dy/dx(1/e^y)=1/e^lnx
Answer: 1/x
Hint: Use properties of Logarithmics
Domain of ln x is (0, positive infinity)
because ln(1/1-x) is greater than 0
By definition y= ln x if and only if x=e^y
e^ln x=x
e^ln cancels
Given y=f(g(x)) then u=g(x) and y=f(u)
g'(x) times f'(u)
u' times f'(u)
Sin^3(x)
(sinx)^3
(cos x)^3 sin^2x
Answer: 3(cos x)sin^2 x
f(u)=x^3
f'(u)=3x^2
u=sin x
u'=cos x
F'(x)=f'(g(x)) times g'(x)
d/dx(cot x) = -csc^2 x
d/dx(sec x)= sec x tan x
d/dx(csc x)= - csc x cot x
d/dx(tan x)=sec^2(x)
d/dx(cos x)=-sin x
d/dx(sin x)=cos x
lim cos x =cos x h→0
lim cos h/h =0 h→0
lim sin x =sin x h→0
lim sin h/h=1 h→0
x^2e^2x
g'=2e^x
2xe^2x+2e^2x(x^2)
Answer 2xe^2x+2x^2e^2x
f'=2x
e=2.71828
d/dxa^kx=klna(a^kx)
d/dxe^x=e^kx
K=real numbers
lime^h-1/h=1 h to 0
(x^2+3x-1)/x^2-1
((2x-3)(x^2-1)-2x(x^2-3x-1)/(x^2-1)^2
Make sure to combine all like terms: 2x^3-3x^2-2x+3-2x^3+6x+2x/(x^2-1)^2
Answer: 4x^2-3/(x^2-1)^2
B=x^2-1
B'=2x
T=x^2-3x-1
T'=2x-3
If f and g are differentiable at x and g(x) does not equal 0, then
d/dx f/g= (f'g-g'f)/g^2 or T'B-B'T/B^2
The derivative of the top times bottom minus the derivative of the bottom times the top all over the bottom squared.
Before starting Product Rule
Assume u=f(x) and v=g(x)
change in v=f(x+change in x)-g(x)
change in u=f(x+change in x)-f(x)
3x^3e^x
g=e^x
g'(x)=e^x
9x^2e^x+e^x(3x^3)
Answer: ex^3e^x+9x^2e^x
f=3x^3
f'(x)=9x^2
(fg)' does not equal f'g'
If f and g are differentiable,
d/dxfg=f'g+g'f
The derivative of the first times the second plus the derivative of the second times the first
f'(x)=lim(f(a+h)+f(a))/h h→0
f'(x)=lim b^x+h-b^x/h h→0
f'(x)=lim b^xb^h-b^x/h h→0
f'(x)=lim b^x(b^h-1)/h h→0
f'(0)f'(x)=limb^h-1 h→0
f(x)=b^x
f'(x)=f'(0)b^x
The rate of change of any exponential function is proportional to the function itself (slope is proportional to height).
if c is a constant then f(x)=C
Process
lim (c-c)/h h→0
lim 0=0 h→0
d/dx(c)=0
b∫af(g(x)g'(x)dx=g(b)∫g(a) f(u)du
1/4u^4+C
1/4(2x+1)^4+C
f(u)=2u^3
u=2x+1
du=2dx
Note: you will need to solve for dx before you can substitute it in.
The antiderivative F(x)+C is the general solution. If we want to find a particular solution, we need to have a point (x,y) of F(x).
Example: F(x)=4x+2 with F(1)=8
∫(4x+2)dx
4(1/1+1)x^1+1+2x+C
2x^2+2x+C
2(1)^2+2(1)+C=8
c=4
F(x)=2x^2+2x+4
dy=∫f(x)dx=F(x)+C
Tells you what you are looking for
So the antiderivative of f(x)=3x^2 is F(x)=x^3+C, where C is a constant
3(1/3+1)x^3+1+3(1/2+1)x^2+1-5(1/1+1)x^1+1-2(1/2)e^2x+C
3(1/4)x^4+3(1/3)x^3-5(1/2)x^2-e^2x+C
Answer: 3/4x^4+x^3-5/2x^2-e^2x+C
simplify
∫b^xdx=1/ln b(b^x)
∫e^axdx=1/ae^ax+C
∫(f(x)+/-g(x))dx=∫f(x)dx +/- ∫(g(x)dx+C
there is an invisible one, when its ∫dx=x
∫x^ndx=1/n+1x^n+1 +C
∫kf(x)dx=k∫f(x)+C
∫kdx=kx+C
∫b^xdx=b^x/ln b +C
∫e^x=e^+C
∫x^ndx=x^n+1/n+1+C
∫cscxcotxdx=-cscx+C
∫csc^2dx=-cotx+C
∫sec^2dx=tan x +C
∫secxtandx=secx+C
∫cosdx=sinx+C
∫sindx=-cos+C
∫1/xdx=ln|x|+C
∫cf(x)dx=c∫f(x)dx
The integral of the rate is the net change
We know f'(x) is the rate of change of y=f(x) with respect to x. Then f(b)-f(a) is the change in y when x goes from a to b. Since y could go up, then down, the up again, between the intervals
Integral
Example: Evaluate the following
∫cos(x/2)dx
1/1/2sin(x/2)+C
Answer: 2sin(x/2)+C
∫cosh(ax)dx=1/asinh(ax)+C
∫sinh(ax)dx=1/acosh(ax)+C
∫1/x√ a^2-x^2dx=sec^-1(x/a)
∫1/√a^2+x^2dx=tan^-1(x/a)
∫1/√ a^2-x^2dx=sin^-1(x/a)
∫csc(ax)cot(ax)dx=-1/acot(ax)dx
∫sec(ax)dx=1/asec(ax)+C
∫csc^2(ax)dx=-1/acot(ax)+C
∫sec^2xax)dx=1/atan(ax)+C
∫sin(ax)dx=1/acos(ax)+C
∫cos(ax)dx=1/asin(ax)
Derivative of trigonometric functions
d/dx(csc ax)=-a csc(ax)cot(ax)
d/dx(cosh(ax)=asinh ax
d/dx(sinh(ax)=acosh ax
d/dx(tan^-1(x/a))=1/a^2+x^2
d/dx(sec^-1(x/a))-1/x√ a^2-x^2
d/dx(sec ax)=a sec(ax)tan(ax)
d/dx(sin^-1(x/a))=-1/√ a^2-x^2
d/dx(cot ax)=acsc^2(ax)
d/dx(tan ax)= asec^2(ax)
d/dx(cos ax)=-a sin(ax)
d/dx(sin ax)= a (cos(ax)
this is a half circle
r=4
A=pir^2(1/2)
1/2pi(2)
2pi
A=lw
(4)(1)
2pi+4
Even and Odd Functions
if f is odd, a∫-af(x)dx=0
If f is even,a∫-af(x)dx=2a∫0f(x)dx
Let a be a real positive number and let f be an integrable function on the interval[-a,a].
Part 2
b∫af(t)dt=f(b)-f(a)
where f is any antiderivative off, that is, a function such that f'=f
plug in a and b into the antiderivative and then subtract them, according to the formula.
Part 1
Derivatives and Integrals are inverses of one another
Let f be continuous for f on[a,b]
g(x)=x∫af(t)dt
is continuous on[a,b] and differentiable on (a,b) and g'(x)=f(x)
Let f be continuous for f to be greater than or equal to a. The area function with constant endpoint a is.
A(x)=x∫af(t)dt
Subtopic
\where x is greater than or equal to a. That gives the net area of the region bounded by the graph of f and the t-axis on the interval [a,x].
Area:b∫a f(x)dx)=limnEk=1(f(x)change of x)
The area function/integral above represents the Net area. Area above the axis is positive, while the area below the axis is negative. Thus, we can get the TOTAL net area of a function.
Integrand-function under the integral
variable of integration- dx
a∫bf(x)dx=-b∫af(x)dx
a∫af(x)=0
with a less than c less then, b∫ak(f(x)+g(x))dx=b∫af(x)dx+b∫ag(x)dx
b∫ak(f(x)+g(x))dx=b∫af(x)dx+b∫ag(x)dx
b∫ak(f(x)dx=kb∫af(x)dx
