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.

About Six Months Ago I figured something out that was very exciting to me. Let me share it with you.
Suppose you have a Certificate of Deposit that is earning a fixed interest rate for a specific term that is compounding daily. Who cares what the term is. Say you want to calculate the exact balance on a specific day. Well I will tell you how you can do it. Of course this will be assuming that it is not a leap year. For leap year calculations just change all of the 365 values in each formula to 366.
Let's first assign variables:
Let P = (The initial amount), r = (The rate), 365 = (365 days in the year), D = (The number of days that our account has been compounding) and I = (The Interest that we will get) and W = (The initial amount + the Interest we get)
The formulas are as follows:

APY (Annual Percentage Yield) = (1+ r/365)^(365) - 1

RATE = ln(APY+1) = [ln(1+r/365)^(365)] = 365*ln(1+r/365) = r

I = P*(1+r/365)^(D) - P So W = (P+I) = P*(1+r/365)^D

D = [ln(W)-ln(P)] / [ln(1+r/365)] where W = (P+I)

So how does this work? Say that we want to know how much interest we would get after 10 days of daily compounding at a 7% rate, with an initial deposit of $100,000.00 to our Certificate of Deposit. Then
I = P*(1+r/365)^(D) - P
So I = $100,000.00*(1+.07/365)^(10) - $100,000.00 = $191.95

Now what if we wanted to know how many days it would take to double our money
Doubling our money would mean that we would have to earn $100,000.00 in interest.
This means we want to know how many days will it take with an initial deposit of $100,000.00 at daily compounding at a rate of 7% to grow to be $200,000.00
So we have D = ln(W)-ln(P)/ln(1+r/365) so we have
ln($200,000.00)-ln($100,000.00)/[ln(1+.07/365)] = 3614.61 days. This means that it would take 3614.61 days to double our money at a rate of 7%

We know that the annual percentage yield is the actual rate of return on an initial deposit that has been compounding at a fixed rate for a full year.

But what if we want to calculate the actual rate of return on any given day after we make the initial deposit?
Well that is easy too. We can take calculate the interest up until that day, and then divide it by our principle account. For example we calculated that we after 10 days on a deposit of $100,000.00 we received 191.95 in interest. Therefore if we wanted to calculate the actual rate of return we would be = to interest / principle = 191.95/100,000 = .0019195 or .19% As you can see although we may be getting 7% as our rate, this rate is actually a yearly rate, in other words, our rate of return actually starts at 0%

Water is leaking out of an inverted conical tank at a rate of 10,000 cm^3/min at the same time that water is being pumped into the tank at a constant rate. The tank has a height of 6m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank?

Hmm… since we are dealing with a conical shape the Volume (V) should = 1/3*pi*r^2*h
And the Area should = pi*r*sqr(r^2+h^2)

We are given dh/dt = 20 dV/dt = C – 10,000 where V = 1/3*pi*r^2*h at time t
A picture will help us figure this out, review my picture below:



r = 2 m
let h = 2 m
height = 6 m
The Diameter is 4 m
1) We have V = 1/3*pi*r^2*h
2) By similar triangles, h/6 = r/2 so
h = r/2 * 6 = 6r/2 = 3r and
r = 2h/6 = 1/3h
4) So we have V = 1/3*pi*(1/3*h)^(2)*h = 1/3*pi*(1/3*h)^3 = pi/27*h^3
5) So dV/dt = pi/9*h^2 *dh/dt
We were given that
6) dV/dt = pi/9*h^2 *dh/dt
We are given that dh/dt = 20
7) So we have dV/dt = C – 10,000 = pi/9*h^2*dh/dt
8) All of these figures are in cm so where as our h value = 2 is in meters so we must convert 2 meters to centimeters = 200 cm
9) Now we can plug in our values h = 200,and dh/dt = 20 to solve for C
C-10,000 = pi/9*(200)^2*(20)
C =10,000 + pi/9*(200)^2*(20) = 289252.68 cm^3/min
Therefore the constant rate at which the water is pored in to the tank is 289,252.68cm^3/min

Todays Math Problem (RELATED RATES)

Boyles’ Law states that when a sample of gas is compressed at a constant temperature, the pressure P and the volume V satisfy the equation PV = C, where C is a constant. Suppose that at a certain instant the volume is 600 cm^3, the pressure is 150 kPa, and the pressure is increasing at a rate of 20 kPa/min. At what rate is the volume decreasing at this instant?

Hmm… we are given that PV = C, where C is a constant, So we can plug our values in for P and V where P = 150 kPa and V = 600 cm^3 therefore we get that 150*600 = C
But we are not asked to find what C equals we are asked to find out how fast the rate of V is decreasing over time. When dP/dt = 20 kPa/min
To figure this one out we need to take the derivatives of our equations using the rules for related rates. Which say that dV/dP = dV/dt*dt/dP so dV/dt = dV/dP*dP/dt and dP/dt = dP/dV*dV/dt so we get dV/dt*dP/dt = dV/dP*dP/dt*dP/dV*dV/dt = dV/dt * dP/dt so since everything else cancels out we will us dV/dt which reads the derivative of Volume in respect to time, and dP/dt which reads the derivative of Pressure in respect to time.

So lets see what we can find out.

Differentiating both sides of the equation PV = C we get P*dV/dt + V*dP/dt = 0 (by the product rule)
Subtracting V*dP/dt from both sides we get P*dV/dt = -V*dP/dt
Dividing P from both sides we get dV/dt = -V/P*dP/dt

Now we can plug in our known values, V = 600 P = 150 and dP/dt = 20
To get dV/dt = -600/150*(20) = -80

Remembering that we are asked to find out what the instantaneous rate of change is which is (dV/dt) when the instantaneous rate of change of pressure is 20 kPa/min which is (Pv/dt) and knowing that the Volume is cm^3 we get a final answer that tells us that as the pressure approaches 20 kPa/min the volume be decreasing at 80 cm^3/min.

Evaluating limits

Some formulas are difficult to evaluate their limits, however, using L’Hospital’s Rule which says that with functions in the form of. 0/0 , (infinity)/(infinity) 0^0, 1^infinity.
We can simply simplify the quotient and take the derivative until our value will work in the remaining function.

Lets take a look at some examples:

Example 1: Suppose we want to take the limit of (x^2-16)/(x-4) as x approaches 4.
If we simply plug in 4 into the function we get (4^2-16)/(4-4) = 0/0
So L’Hospital’s Rule says that we can take the derivative d/dx of the numerator and the derivative of the denominator so lets try that:
Taking d/dx(x^2-16)/d/dx(x-4) we get (2x)/1
Now we may plug in 4 into our new function to get that the limit as x approaches 4 or (x^2-16)/(x-4) = (2)*(4)/1 = 8

Lets take a look at another example:

Example 2: Suppose we want to take the limit of (4x^2-5x)/(1-3x^2) as x approaches infinity
The problem here is that infinity is not a number therefore we can not plug it into the original formula. So lets try take the derivative of the numerator and the derivative of the denominator to see what we yield.
d/dx(4x^2-5x)/d/dx(1-3x^2) = (8x-5)/(6x) we still cannot plug anything into this equations so we will again take the derivate of the numerator and the denominator of
(8x-5)/(6x), lets see d/dx(8x)/d/dx(6x) = 8/6 = 4/3 giving us that the limit as x approaches infinity of (4x^2-5x)/(1-3x^2) is 4/3

Lets do more

Example 3: Suppose we are ask to find the limit of sin(x)/x as x approaches 0 if we plugged 0 into sin(x)/x we get 0/0. so lets take the derivative of the numerator and the denominator of sin(x)/x Doing so we get d/dx(sin(x))/d/dx(x) = cos(x)/1
Now we may plug in 0 into cos(x)/1 to cos(0)/1 = 1/1
So the limits as x approaches 0 of sin(x)/x = 1

Lets try a trickier one:

Example 4: Suppose that we want to find the limit of e^x/x^2 as x approaches affinity. If x is approaching positive infinity then we have that the numerator and the denominator are both approaching affinity. Lets try to apply L’Hospital’s Rule and take the derivative of the numerator and of the denominator and see what happens.
Taking d/dx(e^x)/d/dx(x^2) we get (e^x)/(2x) so this tell us that e^x is increasing faster than 2x so this limit should be infinity, lets check by applying L’Hospital’s Rule again to (e^x)/(2x)
We get that d/dx(e^x)/d/dx(2x) = e^x/2 Now we can see that as e^x increases then 2 will go into it more times to give us the limit as x approaches infinity of e^x/x^2 = infinity

Example 5: Suppose that we want to find the limit of (5t^4-4t^2-1)/(10-t-9t^3) as x approaches 1
More numbers but don’t worry with what we know those will all go away.
Lets apply L’Hospital’s Rule and take the derivative of the numerator and the dominator multiple times to get that d/dx(5t^4-4t^2-1)/d/dx(10-t-9t^3) = (20t^3-8t)/(-1-27t^2)
we could plug in 1 into (20t^3-8t)/(-1-27t^2)
To get 20-8/(-1-27) = 12/(-28) = -6/14 = -3/7 so the limit of (5t^4-4t^2-1)/(10-t-9t^3) as x approaches 1 = -3/7

Lets try applying L’Hospital’s Rule other types of limits:

Example 6: Suppose that we want to find the limit of xlnx as x approaches 0 This function is not in quotient form but we can put it in quotient form xlnx = (lnx)/(1/x)
Now lets take the derivative the numerator and the denominator to get
d/dx(lnx)/d/dx(1/x) = (1/x)/((x*0)-1(1)/(x^2)) simplifying we get (1/x)/(-1/x^2) = x^2/-x = (-x) now we can plug in 0 into (-x) to get (-0) = 0
Therefore the limit as x approaches 0 of xlnx = 0

Lets try this one:

Example 7: Find the limit of xe^x as x approaches negative infinity
Since this is not in quotient form we can apply L’Hospital’s Rule until it is, so lets put it in quotient form. xe^x = (e^x)/(1/x) or xe^x = (x)/(1/e^x)
Now lets apply L’Hospital’s Rule on (e^x)/(1/x) d/dx(e^x)/d/dx(1/x) = (e^x)/(-1/x^2) so that is not going to help us. What if we apply L’Hospital’s Rule to (x)/(1/e^x) then we get d/dx(x)/d/dx(1/e^x) = 1/(e^x*0)-(e^x*1)/(e^x)^2 = (1)/(-e^x)/(e^x)^2 simplifying we get 1/(-e^x) therefore as x approaches –infinity the denominator is getting bigger and which means that 1/(-e^x) will go to 0, so the limit as x approaches negative infinity
of xe^x = 0

Ok one more

Example 8: This one is tricky! Find the limit of x^(1/x) as x approaches infinity it looks like it will be zero but don’t let the eyes deceive you because as x if x = infinity then (1/x) will go to zero witch means that we will have (infinity)^0 which any number x^0 = 1 so the limit of x^(1/x) should be 1
Lets check:
Let y = x(1/x)
Using our friend (ln) we can attach natural log (ln) to both side of the equation to give us
lny = lnx^(1/x) so lny = 1/x(lnx)
now we can see that as x get larger then 1/x will go to 0 which would make 1/x(lnx) = 0
however because we attached natural log to y we must get rid of the natural log by raising the equation lny = 1/x(lnx) by (e) so lets do that. Since we found that as x goes to infinity in 1/x(lnx) then 1/x(lnx) = 0 we can say e^(lny) = e^0 giving us y = 1
Therefore the limit as x approaches infinity of x^(1/x) = 1