If a person starts plucking hair from his head, the person is not bald after plucking first hair, nor after plucking second hair. Then exactly when we can say that person is starting to get bald?

Aristotelian logic i.e the binary logic defines everything in TRUE or FALSE, BLACK or WHITE manner. An element either belongs to the set A or it does not belong to set A, i.e A or not-A, it can not be both. A statement can only be true or false, there is nothing in between.

If we observe the world around us, there is vagueness, there is ambiguity. Things can not always be classified as Black or White but most of the times they have shades.Most of the times human beings think and communicate in gradation and not in crisp manner. In our thought process or the natural languages, meaning of the words is often vague and even when it is well defined, while using the word as a label for a set, the boundaries within which objects belong to or do not belong to a set are vague. A person X does not totally dislike strawberry ice cream, but is not very fond of it either, so she likes it to some degree. Some kids like school very much, some kids completely dislike it, but there are many kids who like some aspects of it and dislike some. Person with age 90 years, is old and person with age 20 years is young, but what about the person with age 35? Is he young or old?

Aristotelian logic falters in the scenario, when terms are not precisely defined. This is where Fuzzy Logic comes into picture.Fuzzy logic defines the terms in more-or-less manner instead of yes-or-no manner.

If we wish to define the set A of tall persons, then we would set a criterion to decide if a person is tall or not. Suppose we decide a person with height 5 feet 9 inches or more belongs to this set then a person with height 5 feet 8 inches does not belong to this set neither does person with height 4 feet 1 inch. So, both these people belong to set not-A.

But does that mean the person having height 5 feet 9 inches is tall and person with height 5 feet 8 inch is not tall? Instead of this crisp definition of sets, Fuzzy logic defines the concept of membership. Every element has a membership in the set, taking value between 0 and 1. The person having height 6 feet may have membership value 1, where as the person having height 5 feet 8 inches, may have 0.8 as membership value to this set and so on. So every element belongs to set A and set not-A as well, to some degree. A piece of cloth can not always be either dry or wet, it can be dry to some extent and wet to some extent, weather can be cold to some degree and hot to some degree.

In 1964, Professor Lotfi Zadeh developed the theory of Fuzzy sets. He was the first one to give this theory a definitive form, though many thinkers in the past had pondered on this question of non-crisp sets. Buddha, founder of Buddhism, who lived in India around 500 BC, says that world is full of contradictions and almost everything contains some of its opposite, i.e things can be A or not-A at the same time. Bertrand Russell had encountered this idea of fuzziness when he stated his paradox.

Today, Fuzzy logic is used in many day-to-day appliances, since it allows more human-like interpretation and reasoning in machines by considering intermediate values between True/False, hot/cold, bright/dark, clean/dirty etc.

Fuzzy washing machines decide the washing cycle required to wash the laundry depending upon its degree of dirtiness. If it is 100% dirty, then say 2 minutes will be added to the basic washing cycle and 1 minute if it is 50% dirty and so on. It can also decide the amount of detergent to be added depending on the above degree of dirtiness of the laundry.

Smart Air-conditioners use fuzzy logic to adjust flow of air according to degree of hotness or coldness. Smart TVs adjust the colour and contrast modes. Japan is the leader when it comes to making use of Fuzzy Logic in order to make systems more intelligent.

In future, more and more systems can be built using Fuzzy Logic which incorporate human thinking, experience and reasoning instead of limiting it to only true/false values.

The great mathematician Kronecker apparently said “God made the integers, all else is the work if Man”. Since then mathematician have managed to build the integers from nothing. Well not quite nothing, but from the empty set, a set containing nothing, and a few rules. These rules prove enough to generate integers, rational fractions, real numbers, infinitesimals and infinite Integers. Fascinating though these are they are not, event he infinite integers, the main concern here. The main concern is what might be called “Classical” infinity, the theory largely developed by Cantor and which (probably incorrectly) is alleged to have driven him insane and to the study of Theology.

The lowest order of infinity

Start with the natural numbers

I= {1,2,3……..} going from one upwards. For every number there is a next number.

Now include the negative numbers and for completeness add Zero. Both negative numbers and zero were oppose by mathematicians of the day and are now accepted, though the notation for negative numbers, and even fractions has changed over the centuries. We get the integers proceeding from zero in each direction.

N = {………-2,-1,0,1,2,3…….}

Comparing infinities

Are there more integers than natural numbers? One of Cantor’s insights was that two sets are equal if the elements of one can be paired off with the elements of the other. Using –> to indicate pairing

0 –>1

1–> -1

2–> 2

3–> -2

and so on. There are as many integers as natural numbers.

How about the even numbers? We can pair each even number with half its value. 2 pairs with one and 4 with two. Zero, being contrary as always, pairs with itself (0/2 =0 after all).

A bigger infinity

Rational fractions can also be paired off with the integers. When we get to real numbers things get interesting. The real numbers, which include the irrationals and the aristocratic crowd of transcendental numbers, cannot be paired off with the integers. The proof is delightfully easy, Assume you have paired off all the real numbers with the integers, thus listing them all, and then construct a real number unpaired with any integer. Turn any real number into a number between 0 and one by putting a decimal point in front of it Making this mathematically precise would take too long). Now take the first number, paired with one, and start your new number with a different digit. Then take the second number and add a digit different from its second digit, and so on, You will end with a number that differs from every listed number in one place. This is called a diagonal argument.

If we say A0 is the infinity given by the size of the set of Integers, we have a second infinity given by the size of the set of real numbers. It does not stop there. It is possible to extend the set of real numbers between 0 and 1 to include the hyper-reals, smaller than any real number but not zero, and then the hyper-hyper reals.. Infinity seems to be breeding.

Infinite types of infinity?

Go back to the integers. It is relatively show the set of all subsets of a set ( known as its power set) is larger than the set even if the set is infinite. We can use this principle to create an infinite number of infinities each larger than the other. But are these the only possible infinities? The answer is undecideable.

Cantor named the infinity that corresponds to the size of the set of real numbers “c”. Mathematical typesetting is not possible there so we use P(X) for the power set of a set X and |X| for its size . Cantor’s continuum hypothesis is that |P(N)|=c that is that every real number corresponds to a subset of the integers. In 1963 this was finally shown to be undecidable though Gödel and Cohen, who between them completed the proof over a period of more than 20 years (but did not formally work together) apparently believed that if the theory of sets is properly extended the Continuum hypothesis will turn out to be false in the extended theory.

The Continuum hypothesis basically says c, the power of the continuum, is the next infinity after the integers. But there may be an arbitrary or even infinite number of integers between the two. No one has yet managed to find even an upper limit for what c might be. P(N) could be larger or smaller than c and even if the continuum hypothesis were true there may be other infinities not reachable by constructing power sets starting with the integers.

Speculation

Gödel believed mathematics describes reality, which is where we started and notes that if the Continuum hypothesis is undecidable then the description of reality is incomplete for in the real world the hypothesis must be true or false. In other words Gödel seems to have believed number has a reality outside the human mind. Investigating that assumption leads into the field of archetypes and Jungian Psychology and is a quest for another day..

Letter Representations:

a, b, c = any x value  on a function.

ƒ(a), ƒ(b) = any output of the respective a or b values in the function

Intermediate Value Theorem

If any function ƒ(x) is continuous on the closed interval [a,b], then there exists at least one c between a and b such that the ƒ(c) is in between ƒ(a) and ƒ(b).

That’s a confusing explanation, but let’s look at it graphically:

You can see here that the function has two x endpoints a and b.  The corresponding y-values (ƒ(a) and ƒ(b)) are also there.  All the Intermediate Value Theorem says is there is at least one point, c, between a and b such that the y-value of c (ƒ(c)) is in between ƒ(a) and ƒ(b).

So if a = 2 and b = 7, the there has to be a x-value, c, in between 2 and 7 on the graph such that the ƒ(c) is in between the ƒ(2) and the ƒ(7).

Put even more simply:  If the thermometer read 50 degrees one hour ago, and it now reads 55 degrees, there must have been a time in the last hour that the thermometer read 52 degrees (or 53, or 51.23, etc).  The point is, the graph didn’t jump straight from 50 to 55 degrees, it had to go up through all the numbers in between 50 and 55 degrees.

Hope this helped!

Π = pi

V = volume

h = height or depth

r = radius

Unlike derivative rules, there is no set rule for a related rates problem.  Instead, you have to look at the information you are given and decide what you are trying to find out.   However, there is a general pattern to solving a related rates problem (I will bold the important steps)  Let’s take a look at this problem:

Example 1:     Air is being pumped into a spherical balloon at a rate of 5 cm3/min.  Determine the rate at which the radius of the balloon is increasing when the diameter of the balloon is 20 cm.

First thing you need to know about related rates is that they are all related to time.  So the derivative will be taken with respect to time.

Solution

First, draw a sketch of what is going on:

First of all, to solve anything in math, we need an equation that relates all of this information.  We know the balloon is spherical, and we are trying to find out what the rate of change (aka derivative) of the radius of the balloon is when the diameter is 20 cm. 

Next, we need to know how to find the volume of a sphere.

Volume of a sphere is   V = (4/3)Πr³

Now list everything you know from reading the problem

V’ = 5 cm³/min (V’ = rate of change of the volume).

r  = 10cm

Now list what you want to know

r’ (the rate of change of the radius

Now, you might be tempted to find out the volume of the balloon right now and plug it in.  But if you did that, there would be no way to find out what r’ is.

This is where implicit differentiation comes in.  Take the derivative of both sides, treating V and r both as variables.

You get:

V’ = (4/3)Π3r²r’

Clean it up to get

V’ = 4Πr²r’

Now remember what we know.  We know V’, and we know r.  We are trying to find out r’.  Since you know 2 out of the three variables, just plug what you know and solve for r’.  Here it is:

(I leave out the units while doing the math, but it doesn’t matter as long as you remember to put them back in the end)

5 = 4Π(10)²r’

5 = 400Πr’

1/(80Π) = r’

So the rate of change of the radius of the balloon when the diameter is 20 is 1/(80Π) cm/min.

This may seem confusing at first, but after you do a couple of these it starts clicking.  You will see there is a general pattern to solving these problems (I will bold the essential steps)

Let’s do another, harder problem.

Example 2:   A tank of water in the shape of a cone is leaking water at a constant rate of 2 ft³/min.  The base radius of the tank is 5 ft and the height of the tank is 14 ft. 

(a) At what rate is the depth of the water in the tank changing when the depth of the water is 6 ft?

(b) At what rate is the radius of the top of the water in the tank changing when the depth of the water is 6 ft?

Solution

Draw a picture of what’s happening.

Knowing this is a cone, we also need the formula for the Volume:

V = (1/3)Πr²h

Next  list everything that we know from the problem:

Total h = 14ft

Radius at the base = 5

V’ = -2 ft³/min

Now list what we want to know:

a) What is h’  when h = 6?

b) What is r’ when h = 6?

These are two individual problems, so let’s deal with them one at a time.

You can see that if we just went ahead right now and took the implicit derivative of the volume formula, we would end up with V’, r’, and h’.  This is too many variables to solve.  So what we need to do is get rid of one of the variables.

Let’s take a look at that picture again.

Remember, the radius of a cone is always proportional to its height.  So looking at this we can see that

r/h = 5/14

We are trying to find h’ in problem a), so it’s a good idea to get everything in terms of h first.

We know that  r = (5h)/14.  Now plug (5h)/14 in wherever there is an r.

Original Equation:

V = (1/3)Πr²h

Modified Equation:

V = (1/3)Π(5h/14)²h

Go ahead and multiply everything out and collect like terms to get:

V = (25/588)Πh³

Now take the derivative of both sides to get:

V’ = (25/588)Π3h²h’

Clean it up to get:

V’ = (25/196)Πh²h’

Now we know V’ and h, so plug them in:

-2 = (25/196)Π(6)²h’

Now just collect terms and the like and solve for h’ (a calculator will be handy):

-2 = (225/49)Πh’

-98/(225Π) = h’

So the height of the water in the cone is dropping at a rate of -98/(225Π)  ft/min (about -.139 ft/min)

Now solve part b)

Now we need to know r’ instead of h’ when h = 6.  Go back to the proportional relationship of the cone:

r/h = 5/14

From this we get that h = (14r/5)

Plug this in to the formula:

V = (1/3)Πr²(14r/5)

Collect like terms:

V = (14/15)Πr³

Take the derivative of both sides and collect like terms:

V’ = (14/15)Π3r²r’ = (14/5)Πr²r’

Now we aren’t given what exactly r is at this point.  However, remember that r = 5h/14 (from our proportional relationship of a cone r/h = 5/14).  So plug 6 in for h (because at this point in time when we take the derivative the depth of the water is 6 ft) to get r = 30/14 = 15/7.

Now that we know both V’ and r, plug them in:

-2 = (14/5)Π(15/7)²r’

Multiply out and collect like terms:

-2 = (90/7)Πr’

Solve for r’ to get:

-7/(45Π) = r’

So when the depth of the water is 6 ft, the radius is shrinking at a rate of -7/(45Π) ft/min  (about .0495)

That’s if for related rates!

Hope this helped!

Letter representations:

√(—–) = square root of what’s inside the parenthesis

c =  any constant

u and v =  any function

^n = any power of a variable.  So x² is the same thing as x^2.  The program I’m writing with won’t let me superscript anything higher than 3, so x^n is the best I can do.

‘ = first derivative (so u’ would mean the first derivative of function u)

Implicit Differentiation:  Most of the time in calculus we have nice y = x formulas.  However, what if we had something like this?

x³ + y³ = 6

Would you want to solve for y and then differentiate the cube root function?  You could, but  I wouldn’t.  However, there is an easier way.  You use implicit differentiation when either solving for y is too much of a pain or when you can’t solve for y.  Here’s how you implicitly differentiate.

Remember, y is the same thing as ƒ(x).  When we implicitly differentiate, we put y’ instead of ƒ’(x).

Here’s the problem.  Treat y like a variable, yet when you find the derivative of y, just put a y’ instead.  Differentiate both sides of the equation.

x³ + y³ = 6

3x² + 3y²y’ = 0   (Take the derivative of both sides!)

The x³ is differentiated like normal.  The y³, however, is not just a variable.  Since y is standing for ƒ(x), that means we have a different function inside of another function, necessitating the use of the chain rule.  So the first derivative of y³ is 3y²(like a normal variable) times y’.

Now that we have differentiated, solve for y’

y’ =     – 3x²   =  -x²

            3y²          y²

Now what?  We have both an x and a y.  What do we do?  This gets to the reason behind implicit differentiation.  Suppose you have this equation:

x² + y² = 4

It’s a circle with radius 2, right?  Now let’s say we wanted to find the derivative at point x = 1.  However, when you think about it, there are two derivatives at that point because the circle has two points existing at x = 1.  If we had tried to differentiate normally, this is what would have happened.

Solve for y:

y = ± √(4 − x²)

Here’s the problem.  We have a ±.  Which one do we choose?  Normal differentiation can’t answer that, so that’s where implicit differentiation comes in.  Implicitly derived:

2x + 2yy’ = 0

y’ =   -x

          y

Now you have the derivative at this point.

Implicit differentiation is confusing sometimes because the derivative relies upon both x and y, rather than just on x.  So when finding the derivative at a point on non-functions (like a circle), you have to find the specific point where you want to differentiate.

Last Example:

x²y² = 6

Use the Product Rule:

2xy² + x²2yy’ = 0

y’ =   2xy²

         x²2y

Cancel like terms to get your answer:

y’ =   y

         x

So at point (1, √(6)), the derivative = √(6)

Letter representations:

c or a =  any constant

u,v,x =  any function (x only used to represent functions in the Rules, otherwise it’s just a variable)

^n = any power of a variable.  So x² is the same thing as x^2.  The program I’m writing with won’t let me superscript anything higher than 3, so x^n is the best I can do.

( ‘ or d/dx) = first derivative of a function (so u’ would mean the first derivative of function u, as would d/dx u).

General Exponential Rule

ƒ’ (a^x) = (a^x)(lna)

Example:  ƒ’ (2^x) = (2^x)(ln2)

Explanation:  Take your original function and multiply it by the natural log of whatever the base is.  Don’t forget to use the chain rule if the power is something other than one!  Example:

ƒ’ (4^(2x + 1)) = (4^(2x + 1))(ln(2x + 1))(2)

Exponential Rule with e (special case)

ƒ’ (e^x) = e^x

Explanation:  This one is easy to remember.  If you did the normal general exponential, you would wind up with (e^x)(ln(e)).  ln(e) = 1, so you are left with e^x, so this is just a helpful rule to know.  Don’t forget your chain rule though!  Example:

ƒ’ (e^(3x²+x)) = e^(3x²+x)(6x + 1)

Logarithmic function

ƒ’ loga(x) =       1      

                     (x)(lna)

loga(x) means log base a of x (I can’t subscript any numbers).  When no a is indicated, it means base 10.  The derivative is 1 over your function times the natural log of a, multiplied by its derivative if you need to do the chain rule.  Don’t forget your chain rule!

Example:  log5(x² + x) =         1        • (2x + 1)  

                                        (x² + x)(ln5)

Natural Logarithmic Function

ƒ’ lnx =  1

              x

Explanation:  It’s that easy to memorize.  Don’t forget your chain rule though!

Example:  ln(x³ + 2x) =       1        • (3x² + 2)

                                      (x³ + 2x)

Hope this helped!

Letter representations:

• c =  any constant
• ƒ(x) and g(x) =  any function
• ^n = any power of a variable.  So x² is the same thing as x^2.  The program I’m writing with won’t let me superscript anything higher than 3, so x^n is the best I can do.
• ( ‘ or d/dx) = first derivative of a function (so u’, would mean the first derivative of function u, as would d/dx u)

Inverse function derivatives can be confusing when you first start out, but don’t worry!  When I took the AP AB Calculus test, they gave me all the parts I needed, just to see if I remembered the formula.

Remember, the inverse of a function is achieved when you switch x’s and y’s and solve for y.  With this in mind, let’s dive in!

Inverse Function Rule

Let g(x) be the inverse of ƒ(x).  So ƒ^-1(x) = g(x).

ƒ’ g(x) =      1    

               ƒ’(g(x))

Explanation:

Short Way (AP Test Tip):  You will usually be given information like: g(…) = … and ƒ’(…) = … in a word problem. on the AP AB Calculus exam (multiple choice) because it is a timed test to test your knowledge of calculus, not algebraic skills.  Doing this the long way takes way too much time for that kind of test.  Example AP test question:

Let g(x) be the inverse of ƒ(x).  If g(3) = 7.2 and ƒ’(7.2) = 4, what is ƒ’ g(3)?  (Note: there may be other “fluff” like the problem saying “The water tower has a drain rate of …. and the inverse of …blah blah blah.  Ignore that stuff.  Just key in on the word “inverse”, then look for the needed parts.

Plug these into the formula:

                    1        =        1        =     1 

               ƒ’(g(3))          ƒ’(7.2)          4

Easy, right?

Long Way:  Now suppose your teacher is not very nice (or just a good teacher that wants you to learn) and makes you do the entire process by hand, here is what you do.  (Note: my teacher did this, and the equation looked really nasty, but when solved it turned out to equal 2 or something like that.  Don’t be discouraged by a tough-looking problem – it may turn out to be really easy!)  Here’s an easy problem:

Let g(x) be the inverse of ƒ(x).  ƒ(x) = 2x + 1.  What is g’(5)?

First we need the inverse of ƒ(x).  So let’s get it

y = 2x + 1

Interchange x’s and y’s:

x = 2y + 1

Solve for y, then replace y with g(x) to remind yourself that this is the inverse:

g(x) = x − 1

             2

Now use the formula:

     1    

ƒ’(g(5))

The g(5) = 2, so plug that in

     1    

  ƒ’(2)

We need to find ƒ’ now.  ƒ’ = 2.  Wait a second.  There is no x to plug the g(5) = 2 into!  That’s not a problem.  Remember we just found the inverse of a power of 1 equation.  The derivative will not have an x.  So for all values of x, ƒ’ = 2.  Plug that in.

  1  

  2

½ is your answer.  For most of your problems that are given to you the long way, however, ƒ’ will have to be found and a value will have to be plugged in.  The last step there is really easy.  (I would have done it but I forgot that lines have no varying derivative when I made the problem up!  Sorry about that.)

Congratulations!  If you have been reading throughout all my articles on all 27 derivative rules, this was the last one!  My next series of articles will talk about the applications of derivatives, like implicit differentiation and related rates.  Until then, keep plugging and chugging away!

Letter representations:

√(——) = square root of whatever is in the parenthesis

c =  any constant

u and v =  any function

^n = any power of a variable.  So x² is the same thing as x^2.  The program I’m writing with won’t let me superscript anything higher than 3, so x^n is the best I can do.

‘ = first derivative (so u’ would mean the firs derivative of function u)

All inverse trig are notated either with an “arc”, like arcsin, or with a ^-1, Like sin^-1(x).  Either way, it means you need to take the inverse of the trig function, not flip it to the denominator (a common mistake).

Here are the inverse trig rules.  Don’t forget the chain rule if you have a function other than x (like x²) !!!

ArcSine

ƒ’ arcsin(x) =           1      

                         √(1 − x²)

ArcCosine

ƒ’ arccos(x) =        -1      

                        √(1 − x²)

ArcTangent

ƒ’ arctan(x) =        1      

                        (1 + x²)

ArcCotangent

ƒ’ arccot(x) =       -1      

                        (1 + x²)

ArcSecant

ƒ’ arcsec(x) =         1             

                       lxl√(x² − 1)

ArcCosecant

ƒ’ arccsc(x) =         -1         

                       lxl√(x² − 1)

Tips to remembering these:  The Co’s (cos, cot, csc) are just the same as the non Co’s except they are negative.  The secants have an absolute value of x, and the x inside of the square root is positive.  The sine and cosine have a negative x in the square root.  Tangent and cotangent have no square root.  Other than that, you just have to memorize these.

Letter representations:

c =  any constant

ƒ(x) and g(x) and h(x) =  any function

^n = any power of a variable.  So x² is the same thing as x^2.  The program I’m writing with won’t let me superscript anything higher than 3, so x^n is the best I can do.

( ‘ or d/dx) = first derivative of a function (so u’ would mean the first derivative of function u, as would d/dx u)

The Chain Rule

d/dx {ƒ(g(x))} = ƒ’(g(x)) • g’(x)

or

d/dx {ƒ(g(h(x)))} = ƒ’(g(h(x))) • g’(h(x)) • h’(x)

etc…(goes in indefinitely)

Explanation:  When you have an individual function within another function, you have to use the chain rule.  When you only have one function inside another, you take the derivative of the outer function (leaving the inner function untouched), and then multiply it by the derivative of the inner function. 

Here is a example:

ƒ’ {(x³+ 1)²}   =   2(x³ + 1)(3x²)   =   6x^5 + 6x²

Here’s a play by play.  Look at the initial function: ƒ’ {(x³+ 1)²}.  There is a function within a function here.  The inner function is x³ + 1.  The outer function is u².  But there is no u² in that, is there?

Yes there is.  Remember, a variable can represent anything (including an entire function!), not just a number.  If you were to say u = x³ + 1, and replace the x³ + 1 with u , violà!  You have ƒ’ (u²)!  Then you use your power rule on the “u²” and get 2u. 

Now remember that u represents x³ + 1.  So plug that back in and you get  2(x³ + 1).  Remember though, the chain rule requires that you take the derivative of the inner part as well and multiply it by what we just did.

The derivative of x³ + 1 is 3x².  Now multiply it by what we just did and you get:

2(x³ + 1)(3x²) = 6x^5 + 6x²

As proof that the chain rule actually works, foil out the initial function.

You get x^6 + 2x³ + 1.  Derive that, and you get 6x^5 + 6x², proving the chain rule

Let’s practice some more.  Can you spot how inner functions there are?

ƒ’ (sin(2x + 1))³

There are 2 inner functions, three “distinct” functions total.  The innermost inner function is 2x+1, the outermost inner function is sin(x) (x represents 2x + 1).  So, following the chain rule:

ƒ’ (sin(2x + 1))³ = 3(sin(2x + 1))²(cos(2x+1)(2)

You could play around with the algebra and trig to make it nice and pretty, but it’s not required in AP AB Calculus, nor is it needed.  Oftentimes it is best to leave it in its original form.

Ok…you’ve seen some pretty easy chain rules.  Now for a doozy.  This will help you understand how complex inner functions can be as well as help you look for inner functions.

How many inner functions are there?

ƒ’  (sin(x² + (cos(2x³ + 1))²)³)²

Answer: 5

Going from outermost to inner most, here are the “parts”.  u = the variable substitution method for each part.

1.  (u)², with u = sin(x² + (cos(2x³ + 1))²)³

2.  sin(u³), with u = x² + (cos(2x³ + 1))²

3.  x² + u², with u = cos(2x³ + 1)

4.  cos(u), with u = 2x³ + 1

5.  2x³ + 1

I won’t try to derive this.  No problem you should get anywhere in any AB Calculus should be this complicated.  This is just an exercise to help you know how to see inner functions.  Don’t worry if you can’t see all the inner functions immediately; seeing inner functions with ease takes multitudes of practice.  It will become easier in time.

Hope this overview for the chain rule derivative helped, and just post any comments, questions, or concerns down below.

Letter representations:

c =  any constant

u and v =  any function

^n = any power of a variable.  So x² is the same thing as x^2.  The program I’m writing with won’t let me superscript anything higher than 3, so x^n is the best I can do.

‘ = first derivative (so u’ would mean the first derivative of function u)

NOTE:  You can only use the trig rules right now if they only have an x to the power of 1 in them.  If you have something like sinx², then you have to use the chain rule plus the trig rule.  I will teach the chain rule in my next article.

Sine Rule

ƒ’ sinx = cosx

Cosine Rule

ƒ’ cosx = -sinx

Tangent Rule

ƒ’ tanx = sec²x

Cotangent Rule

ƒ’ cotx = -csc²x

Secant Rule

ƒ’ secx = (secx)(tanx)

Cosecant Rule

ƒ’ cscx = -(cscx)(tanx)

There are the six rules.  There is no explanation that I could give to any of them that would not venture into the derivation of each rule.  You just have to simply memorize them.  For applications of these rules, you really must know the Chain Rule (perhaps the most important rule for derivatives), which I will post in my next article.

Chain rule + trig rule example:  ƒ’ sin(x³) = 3x²cosx³