General Differential Calculus Formulas, and Functions

Inverse Trigonometric Functions
arcsine x = sin-1 x = y --> sin y = x
arccos x = cos-1 x = y --> cos y = x
arctan x = tan-1 x = y --> tan y = x

Exponential and Logarithimic Functions
log(a) = y --> a^y = x
ln(x) = log(e)x where x is greater than 0
ln(x) = y --> e^y = x

Cancellation Equations
loga(a^x) = x ln(e^x) = x
a^loga(x) = x e^ln(x) = x

Laws of logarithms
loga(xy) = loga(x) + loga(y)
loga(x/y) = loga(x) - loga(y)
loga(x^r) = rloga(x)

Hyperbolic Functions
sinh(x) = (e^x-e^-x)/2
cosh(x) = (e^x+e^-x)/2
tanh(x) = sinh(x)/cosh(x) = [(e^x-e^-x)/2] / [(e^x+e^-x)/2] = (e^x-e^-x) /(e^x+e^-x)
csch(x) = 1/sinh(x) = 2/(e^x-e^-x)
sech(x) = 1/cosh(x) = 2/(e^x+e^-x)
coth (x) = cosh (x)/sinh(x) = (e^x+e^-x) / (e^x-e^-x)

Inverse Hyperbolic Functions
y = sinh-1(x) --> sinh y = x
y = cosh-1(x) -->cosh y = x
y = tanh-1(x) -->tanh y = x
sinh-1 (x) = ln(x+squrt(x^2+1))
cosh-1 (x) = ln(x+squrt(x^2-1))
tanh-1 (x) = 1/2ln((1+x)/(1-x))

Differentiation Rules (where c is a constant number)
d/dx (c) = 0
Example: d/dx (2) = 0 and d/dx (2x) = 2

d/dx [f(x) + g(x)] = f'(x) + g'(x)
Example: d/dx (x^2 +x^3) = d/dx x^2 + d/dx x^3 = 2x+3x^2

d/dx [f(x) - g(x)] = f'(x) - g'(x)
Example: d/dx (x^2 - x^3) = d/dx x^2 - d/dx x^3 = 2x-3x^2

d/dx [f(x)*g(x)] = f(x)*g'(x) + g(x)*f'(x)
Example: d/dx [(x^2)*(x^3)] = [(x^2)*(3x^2) + (2x)*(x^3)] = 3x^4 + 2x^4 = 5x^4

d/dx [f(x)/g(x)] = [(g(x)*f'(x) - g'(x)*f(x))/(g(x))^2]
Example: d/dx [(x^2)/(x^3)] = [(x^3)*(2x) - (x^2)*(3x^2)/(x^3)^2] = 2x^4 - 3x^4/x^6 =
-x^4/x^6 = -1/x^2

d/dx f(g(x)) = f'(g(x))*g'(x)
Example: d/dx [squrt(x^2 + x^3)] = [1/(spurt(x^2 + x^3)) *(2x+3x^2)] =
[(2x+3x^2)/(squrt(x^2=x^3)]

d/dx [cf(x)] = c*f'(x)
Example: d/dx 2*(x^2) = 2*(2x) = 4x

d/dx (x^n) = n*x^(n-1)
Example: d/dx (x^3) = 3*(x^(3-1)) = 3x^2

Exponential and Logarithmic Functions
d/dx (e^x) = e^x
d/dx (a^x) = a^x*ln(a)
d/dx lnx = 1/x
d/dx (loga(x)) = [1/(a*lnx)]

Trigonometric Functions
d/dx (sin(x)) = cos(x)
d/dx (cos(x)) = -sin(x)
d/dx (tan(x)) = sec^2(x)
d/dx (csc(x)) = -csc(x)cot(x)
d/dx (sec(x)) = sec(x)tan(x)
d/dx (cot(x)) = -csc^2(x)

Inverse Trigonometric Functions
d/dx (sin-1(x)) = 1/(squrt(1-x^2))
Proof: Find d/dx (sin-1(x))
Sin-1(x) = y means sin(y) = x
d/dx sin(y) = d/dx (x)
cos(y)dy/dx = 1
dy/dx = 1/cos(y)
With the Pythagorean Theorem which says cos^2(x) + sin^2(x) = 1 or x^2 + y^2 = r^2
Therefore we can say: cos(x) = squrt(1-sin^2) = squrt(1-x^2)
So we have dy/dx = 1/[squrt(1-x^2)]

d/dx (cos-1(x)) = -1/(squrt(1-x^2))
Proof: find d/dx (cos-1(x))
cos-1(x) = y means cos(y) = x
d/dx cos(y) = d/dx (x)
-sin(y)dy/dx = 1
dy/dx = 1/-sin(y) = -1/sin(y)
With the Pythagorean Theorem which says cos^2(x) + sin^2(x) = 1 or x^2 + y^2 = r^2
Therefore we can say that sin(y) = squrt(1-cos^2(y)) = squrt(1-x^2)
So we have dy/dx = -1/[squrt(1-x^2)]

d/dx (tan-1(x)) = 1/(1+x^2)
Proof: find d/dx (tan-1(x))
(tan-1(x)) = y means tan(y) = x
d/dx (tan(y)) = d/dx (x)
sec^2(y)dy/dx = 1
dy/dx = 1/sec^2(y)
With the Pythagorean Theorem which says that 1 + tan^2(x) = sec^2(x) or 1 + x^2 = sec^2(x)
so we have dy/dx = 1/(1+x^2)

d/dx (csc-1(x)) = -1/[(x)(squrt(x^2-1))]
Proof: find d/dx (csc-1(x))
csc-1(x) = y means csc(y) = x
d/dx csc(y) = d/dx (x)
-cot(y)csc(y)dy/dx = 1
dy/dx = 1/-(cot(y)csc(y)) = -1/cot(y)csc(y)
Since 1+ cot^2(y) = csc^2(y)
Then cot^2(y) = csc^2(y) - 1
cot(y) = (squrt(csc^2(y)-1) = (squrt(x^2-1))
And csc^2(y) = 1 + cot^2(y)
So csc(y) = squrt(1 + cot^2(y)) = squrt(1 + x^2)
Therfore dy/dx = -1/cot(y)csc(y) = -1/[(squrt(x^2-1))*(squrt(1 + x^2))]
= -1/[x(squrt(x^2-1))]

d/dx (sec-1(x)) = 1/[(x)(squrt(x^2-1))]
Proof: find d/dx (sec-1(x))
sec-1(x) = y means sec(y) = x
d/dx sec(y) = d/dx (x)
tan(y)sec(y)dy/dx = 1
dy/dx = 1/(tan(y)sec(y))
Since 1+tan^2(y) = sec^2(y) then sec(y) = (squrt(1+ tan^(y)) = squrt(1+x^2)
and tan(y) = squrt(sec^2(y)-1) = squrt(x^2-1)
the 1/(tan(y)sec(y)) = 1/[squrt(1+x^2)squrt(x^2-1)] = 1/[(x)(squrt(x^2-1))]

d/dx (cot-1(x)) = -1/(1+x^2)
Proof: find d/dx (cot-1(x))
cot-1(x) = y means cot(y) = x
d/dx cot(y) = d/dx (x)
-csc^2(y)dy/dx = 1
dy/dx = 1/-csc^2(y) = -1/csc^2(y)
Since 1 + cot^2(y) = csc^2(y) = (squrt(1+x^2))
Then dy/dx = -1/csc^2(y) = -1/(squrt(1+x^2))

Hyperbolic Functions
d/dx (sinh(x)) = cosh(x)
d/dx (cosh(x)) = sinh(x)
d/dx (tanh(x)) = sec^2(x)
d/dx (csch(x)) = -csch(x)coth(x)
d/dx (sech(x)) = -sech(x)tanh(x)
d/dx (coth(x)) = -csch^2(x)

Inverse Hyperbolic Functions
d/dx (sinh-1(x)) = 1/(squrt(1+x^2))
d/dx (cosh-1(x)) = 1/(squrt(x^2-1))
d/dx (tanh-1(x)) = 1/(1-x^2)
d/dx (csch-1(x)) = -1/[x(squrt(x^2+1)]
d/dx (sech-1(x)) = -1/(x)[(spurt(1-x^2))]
d/dx (coth-1(x)) = 1/(1-x^2)

Implicit differentiation
If we have an equation x² + y² = 25 and we want to find dy/dx (the derivative of y in terms of x) we can use implicit differentiation.

Example
Given: x² + y² = 25 and we want to find the equation to the tangent line at the point (3,4) and we know dy/dx = -x/y
Then d/dx(x²+y²) = d/dx (25)
d/dx x² + d/dx y² = 0
2x + 2y dy/dx = 0
So we have 2x +2y (-3/4) = 0
Since dy/dx = -3/4 which is the slope at point (3,4) we have that the equation of the tangent line = 3x + 4y = 25

Minimizing Volume

Pretend as a company, we need to construct a container with no top that has a surface area of 4π ft² What height and base radius r will minimize the volume of the cylinder? For some reason.

Solution:
First we should draw a picture, The picture below should work!

We want to construct the container such that the Area (which we will call A) = 4π
We have two pieces to this container, the bottom which is a circle, and therefore has an area of πr² and the upper portion of a container which is a rectangle when laid out, so its area is 2πrh, so the total area of the container is A = πr² + 2πrh
Since we were given that we want the A = to 4π then we have: 4π = πr² + 2πrh
4π = πr² + 2πrh
Therefore h = (4π - πr²)/2πr = (2/r – r/2) = (4-r²)/2r
So now that we have h = (4-r²)/2r
We want to find the volume, because for some reason we want to minimize the volume.
We have that the V (Volume) = (The Area of the base)*(The height) so we have
V = πr²h
Substituting h = (4-r²)/2r into V we get V = πr²[(4-r²)/2r] = (4πr²-πr³)/2r
Simplifying we get V = πr – πr²/2
Now we can differentiate V finding dV/dr = π – (πr)
Now we want to find a value of r such that π – (πr) = 0
We have r = -π/-π = 1
So we have h = 1.5
And the minimum Volume = about 4.712 or exactly π*(1.5)

Maximizing Area

Say we want to build a rectangular pen with three parallel partitions using 1000 feet of fencing. And we want to know what dimensions to make it so we can maximize the total area of the pen. How would we do this?
Solution:
First we will draw a picture, let the picture below represent our pen:

Now we assign variables: Let the right side and the left side and the three partitions in the middle equal (x) and let the top and bottom = y

Now we will let P = the parameter and A = Area
We are given that the P = 1000
We also know that P = 5x + 2y
So we have 1000 = 5x+2y
So (1000-5x)/2 = y
We know that the Area = A and A = x*y
So by substituting (1000-5x/)2 = y we get
A = x*(1000-5x/)2 = (1000x -5x²)/2 = 500x – 5x²/2
Since we want to maximize the A we want to find the derivative (dA/dx) and find out what values of x make it = 0
So we have dA/dx = 500-5/2*(2x) = 500-5x
Now we want to know what value of x makes 500-5x = 0
We can see that when x = -500/-5 = 100 then 500-5x = 0
So our five sides of our pen should be 100 ft
And our 2 y sides should = (1000-5x/)2 = (1000-5(100))/2 =(1000-500)/2 = 250 ft
So our maximum Area = 100*250 = 25,000 ft²

Maximizing Volume

Suppose that we want to make a box with a square base from 54 ft² of material. What would be the dimensions that will result in a box with the largest possible volume?

First let us draw a picture of this box:


Now Lets assign variables, Let the sides of the base = x and the height = y
We are give that the A = (Area of the base)+(The Area of all of the sides)
And the Area of a square = x*x, therefore we have A = of the box = x² + (Area of the four sides) = x² + 4(x*y)
So we have A = x² + 4(x*y)
We are given that the Area = 54 ft²
Therefore, we have 54 = x² + 4(x*y)
So y = (54-x²)/4x
We are asked to maximize the Volume, so we will let V = Volume
We know that the Volume = side*side*height so we get V = x²*(y)
We can substitute y = (54-x²)/4x into V to get x²[(54-x²)/4x]
Simplifying we get (54x²-x^4)/4x = (54x/4-x^3/4) = (54x-x³)*1/4

Now we can find dV/dx of ¼(54 -x³) = ¼(54-3x²)
Now we want to find a x value such that 54-3x² = 0
We get x = √(54/3) or about 4.24 ft
So y = about 2.12 ft
This means that the maximum Volume can be about 38.18 ft³

So build the Box…

Applications to Business and Economics

The Average Cost Function à c(x) = C(x)/x
The Average Cost Function represents the cost of producing x units of goods or services.
If there is a absolute minimum we can find it by differentiating c(x) = C(x)/x à dc/dx = [x*C’(x)- C(x)]/x^2
i. So When x*C’(x) = C(x) à then dc/dx = 0
ii. The we have that C’(x) = C(x)/x = c(x)
1. Therefore: If the average cost is a minimum, then marginal cost = average cost

Example:
A company estimates that the cost (in dollars) of production x items is
C(x) = 2600+2x+0.001x^2
The company wants to find the cost, average cost, and marginal cost of producing 1000 items, 2000 items, and 3000 items?
They also want to know what production level will the average cost be lowest, and what is the minimum average cost?

Solution:

So we have that the cost function is c(x) = C(x)/x
We were also given that the company found the cost to be
C(x) = 2600+2x+0.001x^2
Therefore we have --> c(x) = 2600+2x+0.001x^2/x
If we use implicit differentiation we get
i. c’(x) = x(2+0.002x)-1(2600+2x +0.001x^2)/x^2
ii. --> c’(x) = 2x+0.002x^2 -2600-2x-0.001x^2/x^2
iii. --> c’(x) = 0.001x^2-2600/x^2
1. c’(x) = 0 when 0.001x^2 = 2600
2. --> x = squrt(2600/0.001) = 1612.45 x gives the number of items that the company should produce in order to minimize cost
3. --> if x = 1612.45 then, we want c(x) to equal C(x) so we get -->
iv. c(1612.45) = [2600+2(1612.45)+0.001(1612.45)^2]/(1612.45) = $5.22 per item
So we have that if they produce about 1612 items of goods and services then the cost per item will be about $5.22 per item.

However, the company wants to find the cost, average cost, and marginal cost of producing 1000 items, 2000 items, and 3000 items?
Because we already found the function c’(x) = 0.001X^2-2600/x^2 is the marginal cost function, and C(x) the average cost function is c(x) = C(x)/x
so we have the following:

Equations used in this solution:
C(x) = 2600+2x+0.001x^2
C’(x) = 2+0.002x
c(x) = C(x)/x = 2600+2x+0.001x^2/x
c’(x) = x*C’(x) – d/dx(x)*C(x)/x^2 = x*C’(x)-C(x)/x^2 = 0.001x^2-2600/x^2
--> our minimum cost is given when c’(x) = 0 in other words when x*C’(x)-C(x) = 0

See the following chart, notice that when C’(x) = C(x) then that is the minimum point.

Considering Marketing, Pricing, and Revenue Associated with Cost
-->The Revenue function is R(x) = x*P(x)
-->The Price function is P(x) = R(x) – C(x)
-->The marginal Revenue is dR/dx
-->The marginal Profit is dP/dx

In order to maximize profit we look for the critical numbers of P
We want to find the critical numbers of P such that P’(x) = R’(x) – C’(x) = 0 thus, R’(x) = C’(x)
So if the profit is a maximum, then marginal revenue = marginal cost

Example:
Determine the production level that will maximize the profit for a company with the given cost and demand functions.
--> total cost for x units is C(x) = 84+1.26x-0.01x^2+0.00007x^3
--> and p(x) = 3.5-0.01x
We want to find the critical numbers of P such that P’(x) = R’(x) – C’(x) = 0

Solution:
-->The Revenue function is given as R(x) = x*p(x) we are given p(x) = 3.5-0.01x
So R(x) = x*(3.5-0.01x) = 3.5x-0.01x^2
So R’(x) = 3.5-.02x
-->We are given C(x) = 84+1.26x-0.01x^2+0.00007x^3
So C’(x) = 1.26-.02x+.00021x^2
Thus marginal revenue R’(x) is equal to marginal cost C’(x) when
3.5- 0.02x = 1.26 -0.02x+.00021x^2
0 = 1.26-0.02+00021x^2+0.02x-3.5
0 = -2.24+00021x^2
Squrt 2.24/00021 = x
x = 103.27 or about 103
Therefore, a production level of 103 units will maximize the profit.