Implicit Differentiation

Q:  Using implicit differentiation, find dy/dx of


at the point (1,-8/49)


Taking the derivative of the right side of the equation is fairly basic, taking the derivative of the left side, on the other hand, is harder.  The left side involves a quotient rule.  I am going to ignore the right side of the equation for now and just deal with the left:


The quotient rule tells you (in some form / notation or another):

(bottom)*(deriv. of top) – (deriv of bottom)*(top) all divided by (bottom) squared.  [You may have learned this with f’s and g’s and f ‘ and g ‘ — it all says the same thing]


Top = y

Bottom = x + 6y

Derivative of Top = dy/dx = y’ [whatever notation you are using]

Derivative of Bottom = 1 + 6(dy/dx) = 1 + 6y’

So, plug this into the quotient rule to get:

[(x+6y)*y’ – (1+6y’)*y] / (x + 6y)2

That is the derivative of the left side of the equation.  Take the derivative of the whole thing (left and right) and you get:

[(x+6y)*y’ – (1+6y’)*y] / (x + 6y)2 = 7x6

We now have our derivative.  We need to solve for y’ by plugging in the point that was given to us: (1,-8/49)

Yuck.  This is a lot of algebra.  I don’t have time right now to show all the algebra, but when I plug in the point I get the following:

y’ = -55/343

I checked this solution by hand and on a computer, so I am fairly confident — double check it yourself!

Use the area of a cirle to find the diameter

Q:  The area of a circle is 16(pi) cm squared.  What is the diameter of the circle?


First we need to know the area formula for a circle… This is:

A = pi*r^2  [r is the radius]

Since the area is given to us, we can plug that in to the formula to get:

16*pi = pi*r^2

The only variable is r, so we can use algebra to solve for r like so:

Divide both sides by pi (which makes the pi’s cancel out):

16*pi /pi = pi*r^2 /pi

16 = r^2

Now, take the square root of both sides to solve for r:

sqrt(16) = r

4 = r

So, the radius of the circle is 4.  The diameter is 2*radius… So, the diameter is 8 cm.

Find all solutions to a Trig Function

Q:  Find all solutions for X where 0 < X < 360 degrees and 9*cos(X) – 5 = 0.


Okay, the 0 < X < 360 just tells you we are looking for answers in one complete circle (no more, no less).

So, we first need to solve for X like so:

9*cos(X) – 5 = 0

9*cos(X) = 5

cos(X) = 5/9

X = cos-1(5/9)

Plug this into your calculator and you get:

X = 56.25 degrees

This is part of your answer, but not all of it.

So, next, since, from our problem, we saw that:

cos(X) = 5/9

this means that the cosine value is positive.  In what quadrants are the cosine values positive??

Quadrant 1 and Quadrant 4 have positive cosine values.  Therefore, there will be 2 answers (one in each quadrant).

So, X = 56.25 degrees is the answer for the angle in Quadrant 1.

To get the angle in Quadrant 4 that also works, you gotta do 360 – 56.25 = 303.75

Therefore, 303.75 degrees is another answer for X.

What this is saying is that:

cos(56.25)= 5/9 and cos(303.75) = 5/9 also.

So, X = 56.25 and 303.75

Simplifying Trig Functions Using Identities

Q:  Simplify cos4t – sin4t


First thing we gotta notice is that this is a difference of squares.  cos4t is the same as cos2t squared.  So, we need to use what we know about factoring from algebra to factor this:

cos4t – sin4t can be factored like so (this just takes practice to see this and realize it):

(cos2t – sin2t)(cos2t + sin2t)

But, now we can use a trig identity because cos2t + sin2t = 1, so plug that in:

(cos2t – sin2t)(cos2t + sin2t)

(cos2t – sin2t)(1) = (cos2t – sin2t)

Now we gotta notice that even though it is simplified a bunch, we have another identity:

(cos2t – sin2t) = cos(2t)

And now we are as simplified as we get.

Amplitude, Phase Shift and Period

Q:  Consider the function y = -5 cos (2x – .1*pi)

a)  Identify the amplitude

b)  Identify the phase shift

c)  Identify the period


The following formulas / concepts will work for sin and cos graphs:

y = a cos (b*(x – h) ) + k

Note:  a, b, h and k are just numbers that will affect the graph.

The amplitude is |a|

The phase shift if 2*pi / |b|

The phase shift is “h”

The vertical shift is “k”

A big thing to notice:  the “b” value is factored out in this formula.

So…. Let’s start with our actual example: y = -5 cos (2x – .1*pi)

Identify who is a, b, h, and k.  Notice, the b is not factored out, so let’s do that firt:

y = -5 cos (2x – .1*pi)

y = -5 cos (2*(x – .05*pi))

[if you multiply the 2 back through, you get the same as the original]


a = -5

b = 2

h = .05*pi

k = 0 (there is no + k at the end of the problem)

a)  The amplitude is |a| = |-5| = 5

b)  The phase shift is h = .05*pi (the graph is shifted .05*pi units to the right)

c)  The period is 2*pi/|b| = 2*pi/|2| = pi

Graphing a Quadratic Function

Q:  Graph the Quadratic Function:  f(x) = 2x2 – 3x + 1


There are two main methods to do this.  I will do both methods (depending on what you have learned in class, you will want to pick either method 1 or method 2 that I show).

Method 1:  Plotting Points

Because this is a quadratic equation, we know the graph will be in the shape of a parabola.  So, I am going to pick a few values for x, then find the y values.  Then, I will plot the points on the graph to make a parabola.  You can pick any values for x when you do this method.

Let’s pick x = 0.  So, plug x into the equation to find y:

2(0)2 – 3(0) + 1 = 0 – 0 + 1 = 1

Point 1:  When x = 0, y = 1:  (0, 1)

Let’s pick x = 1 now:

2(1)2 – 3(1) + 1= 2(1) – 3 + 1 = 2 – 3 + 1 = 0

Point 2:  When x = 1, y = 0:  (1, 0)

Let’s pick x = -1 now:

2(-1)2 – 3(-1) + 1= 2(1) + 3 + 1 = 2 + 3 + 1 = 6

Point 3:  When x = -1, y = 6:  (-1, 6)

Now, plot the points (0, 1) and (1, 0) and (-1, 6) on a graph.  Connect the dots to make a parabola!

Keep in mind:  You can get more than 3 points for more accuracy.

Method 2:  Finding the y-intercept and the vertex:

The y-intercept of any type of graph occurs when you plug in 0 for x.  So, plug in 0 for x into the equation and solve for y:

2(0)2 – 3(0) + 1 = 0 – 0 + 1 = 1

So, the y-intercept is the point (0, 1).

Now, there is a formula to find the vertex of a quadratic equation.  It goes as follows:

Step 1 to find the vertex:  x = -b / 2a (this will give us the x coordinate of the vertex.

Recall, in our equation, a = 2, b = -3 and c = 1

So, the x part of the vertex is x = -(-3) / 2(2) = 3/4

Now, to get the y part, you need to plug the x part into the equation and solve:

2(3/4)2 – 3(3/4) + 1 = -1/8

The vertex is the point (3/4 , -1/8)

So, on your graph, plot the vertex and the y-intercept.  Draw a parabola through the y-intercept until you hit the vertex.  You know the vertex is either the high or the low point.  Then, “bounce off” the vertex and continue drawing the other half of the parabola.


Simplify by Rationalizing the Denominator

Q:  Simplify by rationalizing the denominator:  √8/√24

Answer:  To rationalize the denominator, multiply both top and bottom by the denominator.  So, multiply both top and bottom by √24:


√8*√24 / √24*√24

Simplify the top and bottom like so:

√192 / 24

Now, we need to simplify the numerator!  It turns out that 192 = 64*3, so:

√192 / 24 = √64*√3 / 24

And, this simplifies to:

8*√3 / 24

Now, divide an 8 out of top and bottom to get:

√3 / 3

Trig Functions and Exact Values

Q:  Let X be an angle in quadrant III such that cos(X) = -12/13.  Find the exact values of csc(X) and cot(X).

This will be a little hard to explain, since I cannot currently provide a picture, but it should be manageable!  Draw a picture on your own paper and follow along with me.

Draw a right triangle and label one of the angles X (just not the right angle).  We are going to ignore the negative sign for now and then take it in to account later!  Since the cos(X) = 12/13, the side adjacent to X is 12 and the hypotenuse is 13.  Use the Pythagorean Theorem to find the missing side (the side opposite X).  You will find that the opposite side is 5.

So, Adj = 12, Opp = 5 and Hyp = 13

Now, let’s consider the negative sign:  Since X is a QIII angle, we know that only tangent and cotangent are positive values in that quadrant — all other trig functions in QIII are negative.

Now, we need to find the csc(X).  “csc” is the ratio of the hyp / opp [and, it will be negative]

So, csc(X) = -13/5

And, cot(x) is the ratio of the adj / opp [and, it will be positive].  So, cot(X) = 12/5


Condensing Fraction (involving variables)

Q:  Condense:  3/2x + x/(2x-6)


In order to add fractions, we need to have common denominators.  Remember, in order to get common denominators, we need to multiply by something.. Just because one denominator is 2x – 6 and the other is 2x, we cannot just subtract 6 to make them match.

So, the fraction on the left needs a (2x-6) and the fraction on the right needs a (2x)…  So, you have to multiply the fraction on the left by (2x-6) and the fraction on the right needs to be multiplied by (2x).  Remember:  multiply to the top and the bottom like so:

3/2x + x/(2x-6)

3(2x-6)/2x(2x-6) + x(2x)/(2x-6)(2x)

Now, simplify each numerator, leave the denominators alone:

6x – 18 / 2x(2x-6) + 2x2 / 2x(2x-6)

Now, since the denominators are the same, we can just combine the numerators to make one fraction.  The denominator goes unchanged:

(6x – 18 + 2x2) / 2x(2x-6)

I will rearrange the numeartor:

(2x2 + 6x – 18) / 2x(2x-6)

Now, let’s factor the top to see what we can cancel out (if we can):

2(x2 + 3x – 9) / 2x(2x-6)

The numerator cannot be factored any more, so we can cancel a 2 out of the top and out of the bottom to get:

(x2 + 3x – 9) / x(2x-6)

This is either the final answer, or you can multiply out the denominator to get:

(x2 + 3x – 9) / (2x2 – 6x)

Solving a quadratic

Q:  Solve for n:

4n2 + 3 = 7n

Answer:  Since there is an n-squared term, this is a quadratic equation.  In order to solve this, we need to set the whole equation equal to 0 first (so, let’s subtract the 7n over to the left side of the equation):

We get:

4n2 – 7n + 3 = 0  [notice that I put the n’s in order of n-squared, n, and then the constant 3]

Now, there are a few methods you may have learned to can help you solve this:  1)  Factoring or 2) Quadratic Formula or 3) Completing the Square.

This is factorable, so I am going to solve it by factoring.  Keep in mind, if you have never factored something this complicated, you will need to plug it into the quadratic formula (no big deal, just use a = 4, b = -7, c = 3 and plug it in to the big ol’ formula you have).

To factor… set up your parenthesis:

(            )*(              )

The first two terms need to multiply to give 4n2 and the back two terms need to multiply to give + 3.  To multiply and get +3 we either need 2 positive numbers or two negative numbers.  Specifically, we need -1 * -3 or 1*3.

Since we have a -7n in the middle, the back two terms both need to be negative.  Therefore, the two back terms must be -1 and -3:

( ___ – 1 )*( ___ – 3)

Now we need to figure out the front terms.  They must multiply to give 4n2 .  This leaves us with either 4n*1 or 1*4n or 2n*2n.

So you can try each combination in place, distribute it out by hand or visually to see which one will “work”.  Factoring does involve a decent amount of guessing and checking if you want to be good at it.

Through a little guessing and checking, I get:

4n2 – 7n + 3 = (n – 1)*(4n – 3) = 0

Which means either:

n – 1 = 0  or  4n – 3 = 0

Solve for n in each case to get:

n = 1 or n = 3/4

You will get the same answers if you had used the quadratic equation to solve (which may be easier if you are not comfortable with factoring more complicated expressions)