ò xⁿ dx = [ x ^{( n + 1 )}/ (n+1)] + C, n ¹ -1

y = x ^{( n )} Þ y ′ = n * (x^{( n + 1 )} )

Ex: y= 3^^{((x^2)+4x)} Þ (dy/dx)= (3^^{((x^2)+4x)} * Ln3) * (2x+4)

Ex: U sub: u = ((x^2)+4x) u’ = (2x+4) for above example.

Implicit diff: (dy/dx) = n* x ^{( -1 )} * y = n*x ^{( -1 )} * x ^{( n )} = n* x^{( n - 1 )} Þ (dy/dx) = n * x^{( n - 1 )}

A Logaritym is an Exponent :

2^^(x) = 5 à Log 2^^(x) = log 5 à x log 2 = log 5 à x= [(Log 5) / (Log 2)] ≈ 2.32

e raised to exponent: (d/dx) [e^^(x) ] = e ^^(x) à Derivative is itself.

That's right. The function f(x) = e^x is its own derivative.

Integrate of e raised to exponent: ò [e^^(x) ] = e ^^(x) + C

===========================================================

Change of Base Formula:

log _b x = log _a x / log _a b Also: log _b x = Ln x / Ln b

Derivative with respect to x: (d/dx) log _b x = 1 / x (Ln b)

y = b^^(x) , b > 0, b ¹ 1 Þ Ln y = Ln b^^(x) = x *(Ln b)

** Diff Formula of b^^(x) Þ (d/dx) b^^(x) = b^^(x)*(Ln b)

===========================================================

Natural Log Stuff:

ò xⁿ dx = Ln | x | + C, n ¹ -1

Ln y = Ln x ^{( n )} Þ Ln y = n Ln x Þ 1/y (dx/dy) = n * (1/x)

Ln e = 1, Ln 1 = 0 ….. 2^x = 5 Þ x Ln 2 = Ln 5 Þ x = (ln 5) / (Ln 2) » 2.3219…

Definition of Natural Log Function: Ln x = ò (upper= x, lower=1) (1/t) dt, x > 0

ò (1/t) dt Þ f(t) = (1/t) Þ f(t) = t ^{( - 1 )} Þ f(t) = t ^{( - 1 + 1 )} / 0 f(t) = undefined.

Ln =ò (upper= x, lower=1) (1/t) dt Þ (dy/dx) [ Ln x } = (dy/dx) ò (upper= x, lower=1) (1/t) dt = 1/x, x > 0

f(x) = Ln (expression) u = expression. f(u) = Ln u Þ f ‘(x) = f ’(u) * u′ >  Þ f ′ (x) = (u’/ u)

f(x) = Ln (x+1)^(1/2) Þ (1/2) Ln (x+1) Þ f ’(x)= (1/2) (d/dx) [ Ln (x+1) ] + Ln (x+1) (d/dx) (1/2)

(dy/dx) [ Ln u ] = (u ′ / u) Þ ò (u ′ / u) dx = (dy/dx) [Ln | u | ] Þ ò (u ′ / u) dx = Ln | u | + C

//=== Proof that f(x) = Ln(x), f ‘(x) = (1/x) ========//

We know from the discussion so far that ln(x) and e^x are inverses of each other.

That means that if f(x) = ln(x), then

 ef(x) = eln(x) = x eq. 6.3-12

Now remember that e^x is its own derivative.

Using the chain rule to take the derivative of the left hand term

(we'll ignore the middle term),

and elementary derivatives to take the derivative of the right hand term,

we have

 ef(x) f'(x)  = 1 eq. 6.3-13a

But from eq. 6.3-12, we know that  e^f(x) = x.

So we substitute x in for that.

 x f'(x) = 1 eq. 6.3-13b

or equivalently

 1
 f'(x) =   eq. 6.3-13c
 x

And that's the derivative of the natural log function -- 1/x.

Remember it. It comes up all the time.

/////==========================================================/////

/////= More on Differential Formulas f(x) = Ln (expression) u = expression. =/////

f(x) = Ln x f ‘ (x) = (1/x)

Example: p(x) = Ln (x^² + cos ^² (x)) let u = (x^² + cos ^² (x))

with u-substitution: p(x) = Ln (u(x))

p ’(x) = (1/(u(x)) * u ‘(x) o r (u ‘(x) / u(x))

p ’(x) = (2x+2cos(x) -(-sin(x))/( (x^² + cos ^² (x)) or (2x-2cos(x)sin(x) )/( (x^² + cos ^² (x))

//=======================================================//

/////= More on Differential Formulas Log(ex-ression) =/////

y= log ₃ (x^⁽²⁾ + 4x) à Deriv of (x^⁽²⁾ + 4x) is (2x + 4) Deriv of log ₃ is (Ln(3)

y ‘ = (2x + 4) / (Ln(3) (x^⁽²⁾ + 4x) )

//=======================================================//

Trig Identities:

ò sin (x) dx = - cos (x) + C ò cos (x) dx = sin (x) + C

ò tan(x) dx = ò (sin(x)/cos(x))dx = - ò( -sin(x))/(cos(x))dx = - Ln |COS X| + C

ò cot(x) dx = ò (cos(x)/sin(x))dx = Ln |sin X| + C

ò sec(u) dx = Ln |sec u + tan u| + C ò csc(u) dx = - Ln |csc u + cot u| + C

================================================================

(d/dx) e^^(u(x)) = (e^^(u(x)) )*(u′(x)) Þ Composition therefore Chain Rule.

///====== Examples Derive of trig function in Ln ========///

y = (sin(x))^^{(x^2)} Þ Ln y = Ln (sin(x))^^{(x^2)} Þ Ln y = (x^2)*(Ln (sin(x)))

(1/y)(dy/dx) = (x^2) * ((cos(x) / (sin(x))) + 2x Ln(sin(x)) Þ

(dy/dx) = [ (x^2) * ((cos(x) / (sin(x))) + 2x Ln(sin(x)) ] * Y Þ

(dy/dx) = [ (x^2) * ((cos(x) / (sin(x))) + 2x Ln(sin(x)) ] * [(sin(x))^^{(x^2)} ]

///====== Examples Derive of a function in Ln =========///

y = [(x-5)^{^(5)} ] * [(x^{^(2)} +4) ^{^(3)} ] * [ (3x + 5) ^{^(2)} ] Þ

** Ln y = Ln [(x-5)^{^(5)} ] * [(x^{^(2)} +4) ^{^(3)} ] * [ (3x + 5) ^{^(2)} ] Þ

** Ln y = (5) Ln [(x-5)] * (3)Ln [(x^{^(2)} +4) ] * (2) Ln [ (3x + 5) ] Þ

(1/y)(dy/dx) = [ (5 * 1)/(x - 5) + (3 * 2x)/( x^{^(2)} +4) + (2 * 3)/ (3x + 5) ] Þ

(dy/dx) = [ (5 )/(x - 5) + (6x)/( x^{^(2)} +4) + (6)/ (3x + 5) ] * Y Þ

(dy/dx) = [ (5 )/(x - 5) + (6x)/( x^{^(2)} +4) + (6)/ (3x + 5) ] * [[(x-5)^{^(5)} ] * [(x^{^(2)} +4) ^{^(3)} ] * [ (3x + 5) ^{^(2)} ]]

///====== Examples Inter to Ln Using U Substitution =========///

ò (4 x e ^{^(x^(2) + 2)}dx Þ Let: u =(x^(2) + 2) then: du = 2x dx (have to make 4 adj)

2 ò (2 x e ^{^(x^(2) + 2)} dx = 2 ò e^{^(u)} du = 2 e^{^(u)} + C = 2 e^{^( x^(2) + 2)} + C

///=== More Examples Inter to Ln Using U Substitution =========///

ò(9x^{^(2)} – 6)/ (3x^{^(3)}-6)dx Þ Let: u =(3x^{^(3)}-6) then: du = (9x^{^(2)} – 6) dx (is deriv of denominator)

ò((du)/(u)) = Ln|(u)|+ c = Ln|(3x^{^(3)}-6)|+ c

Always enclose |u| with absolute symbols for safety with Ln.

///=== More Examples Inter to Ln Using U Substitution =======///

This one has Inter with upper and lower limits that need to adj

The ò upper limits = e. lower limits = 1.

ò (Ln x)/(2x) Since Ln(e) = 1, then u=(Ln x), and du= (1/x) (but it is 2x, multiply by ½)

½ ò (Ln x)/(x) Since u is used, adj limits to match. Upper = 1, lower = 0 Þ

½ ò (u)(du) = ½ [(u^(2)/2] = [(1/4) – 0] = ¼

///=== More Examples Inter to Ln Using U Substitution with trig f(x) =====///

first: tan= (sin/cos) Þ u = cos(x). Þ du = - sin(x) (because neg sin, multiplyò by –1)

ò tan(x)dx = - ò (- sin(x) / cos(x)) = - ò (du/u) = - ln|u| + C = - ln| cos(x)| + C

///=== /////= Integration =//////// ===///

ò e^{^(u)} du = e^{^(u)} + C or Þ ò e^{^(u)} u’ dx = e^{^(u)} + C

other than e Þ 5 = e^{^(Ln5)} , y = 5, Ln y = Ln 5 Þ Therefore y = e^{^(Ln5)}

///===/////= Integration Example=////////===///

ò 2 ^{^(x)} dx Þ 2 = e ^{^(ln2)} Þ or ò 2 ^{^(Ln2) x} dx

Let: u = (Ln2) x and then Let: u ’ = Ln2 ( u’ or du)

= (1/Ln 2) ò 2 ^{^(Ln2) x} Ln2 dx (adjustment was made for 2dx by multiply of (1/Ln 2))

= (1/Ln 2) * e ^{^(Ln2) x} + C

///=== /////= Population =///// ===///

Rate of change of Population is Proportional to Population.

P is for Population, k is for Rate of Change, t is for time. (Rate of change always involves time.)

(dP/ dt) = k * P Þ ò(1/P)dP = ò k dt Þ

Ln P = k * t + C₁ Þ P = e^{^(kt + c}₁⁾ = e^{^(kt )} e^{^( c}₁⁾ = Ce^{^(kt )} .

P = Ce^{^(kt )} .

P₍₀₎ = Ce⁽⁰⁾ is Population at zero time

Use earliest time as your time zero.

//// Example of Population Growth: ////

P₍₀₎ = 100 Million, P₍₂₀₎ = 120 Million, P₍₄₀₎ = ?

P = Ce^{^(kt )} but now we know that C = 100

P = 100e^{^(kt )} Formula to use to solve for k

P at 20 yrs was P₍₂₀₎ = 120 Million

120 = 100 e^{^(k*20 )} ..Now solve for k

(120/100) = e^{^(k*20 )} Þ Ln(6/5)= Ln e^{^(k*20 )} Þ

Since Ln e = 1 then Ln(6/5)= (k*20 ) Þ (Ln(6/5))/ 20 = k

k ≈ 0.009116 now place into formula to get population of any year.

P = 100e ^{^( 0.009116} * ^{t )} Formula to use to solve for population of any year

P₍₄₀₎ =100e ^{^( 0.009116} * ^{(40) )} Solve for P. P ≈144 Million.

**** if asked about how long to double…

then double C and make it the new PÞ

200 = 100e^{^( 0.009116} * ^{t )} Þ 2 = e^{^( 0.009116} * ^{t )} Þ Ln 2 = 0.009116 * t

Now solve for t Þ (Ln 2) / (0.009116) = t Þ t ≈ 76.0357 yrs

answer: It takes about 76 years for the population to double. .

**Note the growth rate Þ (Ln 2) / (0.009116) was for 2 yrs…

Therefore: To triple would be Þ (Ln 3) / (0.009116)

Just change the number after Ln to apply that rate of growth.

///=== /////= Temperature =///// ===///

Newton’s Cooling Law: (dT/ dt)= k(T – T_s)

T₀ is the initial temperature of the object;
C is the temperature of the environment;
k  is the cooling rate constant (constant for a particular material);
t  is the time since T₀; and
e is the exponential function.

** note: as temperature of Medium and Object get closer, cooling is slower.

(dT/ dt)= k(T – T_s) Þ (1/(T – T_s))dT = k dt Þ

Ln|(T – T_s)| = k*t + C₁ Þ

|(T – T_s)|= e^^{(k*t +c}₁⁾ = e^^(k*t) + e^^(c₁⁾ Þ C * e ^^(k*t)

Final formulas:

|(T – T_s)|= C * e ^^(k*t) Þ T_(t) = T_s + C * e ^^(k*t)

//// Example of Change in Temperature: ////

T_s = 44, T_{( 0 )} = 70 T_{( 2 )} = 58, T_{( 9 )} = ???

T_(t) = T_s + C * e ^^(k*t) Þ Since the Surrounding temp is 44 then:

T_(t) = 44 + C * e ^^(k*t) Þ is our formula to work with,,, now solve C.

70 = 44 + C * e ^^(k*t) Þ Subtract 44 to start isolating C.

(26) = C * e ^^(k*t) Þ But time is Zero there for e cancels

C = 26 Þ Now we substitute the C into the problem.

**** T_(t) = 44 + 26 * e ^^(k*t) Þ Now solve for k by using know temp & time.

58 = 44 + 26 * e ^^(k*(2) Þ Subtract 44 to start isolating k.

14 = 26 * e ^^(k*(2) Þ Divide by 26 to get K alone.

(14/26) = e ^^(k*(2) Þ change to Ln so we can get Ln e = 1.

(Ln(14/26)) = Ln e ^^(k*(2) Þ Eliminate the e since Ln e = 1.

(Ln(14/26)) =(k*(2)) Þ Last step is to divide by 2 (time).

(Ln(14/26))/ (2)) =(k)Þ Solve for k. k ≈ -0.30952

T_(t) = 44 + 26 * e ^^{(-0.30952 * t)} Þ Formula to use to solve for any time.

T_{( 9 )} = 44 + 26 * e ^^{(-0.30952 * 9)} Þ Plug and Play T_{( 9 )} ≈ 45.6039

Therefore: After 9 minutes the temperature will be 45.6 degrees.

.