Expert: Scotto - 5/29/2011

Question
Hi, my question isn't necessarily a homework one, but more of an academic circumstance in need of advice. See, I'm currently a full time college student, and I just completed College Algebra and Trig in the same semester to accelerate my advancement into Calculus. I love Math, and I've been a straight A student so far. However, the problem arose when I missed a lot of class due to illness and was forced to cram just to pass without appreciating what I was learning.

So, my question is: should I still go into Calculus (making it my only focus) in the fall knowing I'd have to work much harder to compensate for any gaps, or should I set myself back a semester, enter Pre-Calculus and give my brain a break? I'm an Engineering major who wants to keep things moving, but I just don't want to get burned out again.

On a last note, I have to register this week which doesn't give me much time to make a solid d

Answer
With your determination in taking both trig and college algebra in the same semester,
and you're love of mathematics, I would go for it.

For some major prep work, continue on...

In calculus, you learn about derivatives (which determine the slope)
and integrals (that determine the area).

Derivative
-----------------------------
The derivative of y = f(x) can be written as dy/dx, y', or f'(x).
They all mean the derivative of y.

If y = Cx^n, y' = nCx^(n-1) where is a constant;
example: y = 3x^5' then to find y', multiply by 5 and subtract one from the exponet.
This means y' = 15x^4.

For trig, if y = sin(x), then y' = cos(x);
if y = cos(x), then y' = - sin(x);
if y = tan(x), then y' = sec^2(x).
Example: if y = 5sin(x), then y' = 5cos(x); if y = 6tan(x), y' = 6sec^2(x).

This won't be until much later, but if y = e^x, that is the only function that doesn't change.
What I mean is that if y = e^x, y' = e^x.  The inverse of e^x is ln(x), and if
y = ln(x), y' = 1/x.

Note that any of the preceeding y's could be replaced by f(x),
and y' would be replaced by f'(x).  This becomes important when more variables are involved.
If y = xyz, then y' is meaningless, since we don't know which variable is being used.
If we said f(x) = xyz, then f'(x) = yz, since the power on x was 1, so I multiplied by 1 and reduced the power by 1.  If we said f(y) = xyz, then f'(y) = xz,  If we said f(z) = xyz,
f'(z) = df/dz = xy.


Integral
-----------------------
To integrate, just reverse the process.
For example, if f(x) = 12x^5, the integral is the reverse, so raise the power by 1,
and divide by the new exponet.  The intgral of f(x) is usually referred to as F(x).
So, if f(x) = 12x^5, adding one to the power makes the new power 5+1 = 6.
Dividing 12 by 6 gives a leading 2, so F(x) = 2x^6.

Now on integrals, an are below a curve is found, so an upper and lower limit are provided.
If the upper were 6 and the lower were 3, the value would be 2*6^6 - 2*3^6 =
2(6^6 - 3^6).  To get 6^6, that is 6*6*6*6*6*6.  Well, 6*6 = 36, 36*6 = 216,
216*6 = 1296, 1296*6 = 7776, and 7776*6 = 46,656.
To get 3^6, 3^6 = 3*3*3*3*3*3; 3*3 = 9, 3*9 = 27, 3*27=81, 3*81=243, and 3*243=729.
So the final answer is 2(46,656 - 729) = 2(45,927) = 91,054.

Now if you're really into mathematics, it is kind of interesting to note that
3^6 = 729 and the difference ended in the reverse, 927.



Initially
---------------
To start calculus, know lim(h->0) means "the limit as h approaches 0".

The definition of f'(x) is f'(x) = lim(h->0)[f(x+h)-f(x)]/h.  
As can be seen, this is 0/0, but an x always cancels.

Let's take a simple function, like f(x) = 3x^2.
This means that f(x+h) = 3(x+h)^2.
So we have f'(x) = lim(h->0)[3(x+h)^2 - 3x^2]/h.
Working it out gives (3(x^2 + 2xh + h^2) - 3x^2)/h,
and then multiplying it out gives (3x^2 + 6xh + 3h^2 - 3x^2)/h.
As can be seen, 3x^2 - 3x^3 in the numerator cancel, leaving (6xh+3h^2)/h.
Cancelling the h gives 6x + 3h.
Taking the limit as h goes to 0 gives 6x.

They use this heavily at the start, but it is soon a forgotten topic only to be used when explaining calculus to those who don't know it yet.


Summary
-------------------
In summary, here are the principles:
If f(x) = Cx^n, f'(x) = nCx^(n-1);
if f(x) = C*sin(x), f'(x) = C*cos(x);
if f(x) = C*cos(x), f'(x) = -C*sin(x);
if f(x) = C*tan(x), f'(x) = C*tan^2(x);
if f(x) = Ce^x, f'(x) = Ce^x.
integral(f(x)) = F(x);
integral(x^n) = x^(n+1) / (n+1)  +  C;
integral(sin(x)) = -cos(x) + C;
integral(cos(x)) = sin(x) + C; and
integral(ln(x)) = 1/x.

There will be more, but that is enough info to get started through 4 terms of calculus.
Note that in integrals and derivatives, constant multipliers always carry through and
are multiplied by.  The derivative of a constant by itself is 0.  The integral of a
constant C is Cx if x is the variable.

In most calculus classes, as they get more advanced,
y and f(x) are the function and x is the variable.
For f(x) sometimes they use g(x) or h(x), but any letter can be used.

Feel free during the term to select me when a question is to be asked.