# Math Help - Calculus Notes

1. Originally Posted by janvdl
You may want to add that a simple method to find for example the 125th derivative of sin x would be to divide 125 by 4 repeatedly until you get a remainder smaller or equal to 4.
Not repeatedly, sorry about that! You should subtract 4 repeatedly or else just divide 4 into it, then work with the remainder.

2. ## Lesson 3.4 - Deriving Exponential Functions

Lesson 3.4 - Deriving Exponential Functions
Exponential functions are any function whose power is the variable and base is a constant.

$f=a^x$

The derivative ALWAYS follows this form.

$f'=a^x ln|a| x'$

Functions of base e have the nice property of having the natural log disappear...

$f=e^x$
$f'=e^x ln|x| x'$
$f'=e^x$

Example 3.4-A
$f=2^t$
$f'(t)=2^t ln|2|$

What about taking the derivative of a log?

$g(t)=ln(t)$
$g'(t)=\frac{1}{t}$

The rule is to take one over what is being "logged". If the log is not of base e, then the natural log of that base needs to be divided as well.

Example 3.4-B
$R(w)=log_{10}(w)$
$R'(w)=\frac{1}{ln(10)w}$

Now, what is e?

As a number, it is equal to $2.71828182845905...$ Where does this number come from? It actually has a few definitions:
1) $e = \lim_{n \rightarrow \infty} (1 + \frac{1}{n})^n$
2) e is the unique positive number for which $\lim_{h \rightarrow 0} \frac{e^h-1}{h} = 1$
3) $e = \sum_0^{\infty} \frac{1}{n!}$

The second definition can be easily shown using the definition of a derivative.

Example 3.4-C
$f(x)=e^x$

$f'(x) = ?$

$f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$f'(x) = \lim_{h \rightarrow 0} \frac{e^{x+h} - e^x}{h}$
$f'(x) = \lim_{h \rightarrow 0} \frac{e^x(e^h - 1)}{h}$
$f'(x) = e^x \lim_{h \rightarrow 0} \frac{e^h - 1}{h}$
$f'(x) = e^x$

Not much further detail can be discussed until the all mighty chain rule is introduced. Stay tuned until the next lesson.

3. Originally Posted by colby2152
$f=a^x$

The derivative ALWAYS follows this form.

$f'=a^x ln|a| x'$
Just a small point: you don't need the absolute value signs round the a, because the function a^x is only defined when a>0.

An even smaller point: \ln looks better than ln in TeX, also \log for log. Same applies to \cos etc. for trig functions.

4. Originally Posted by Opalg
An even smaller point: \ln looks better than ln in TeX, also \log for log. Same applies to \cos etc. for trig functions.
even smaller point: \ln, \log, \lim etc also function better, as in, when they have indexes and such.

case in point: $$\lim_{x \to \infty} f(x)$$ yields $\lim_{x \to \infty} f(x)$

while $$lim_{x \to \infty} f(x)$$ yields $lim_{x \to \infty} f(x)$

well, it's still all about the looks

5. ## Lesson 3.5 - Chain Rule

Lesson 3.5 - Chain Rule
The chain rule is one of the most important derivative rules that there is.

Say we have a composition of a function such as:

$h(x) = g[f(x)]$

The derivative of h with respect to x is the derivative of g multipled by the derivative of what's INSIDE g...

$h'(x) = g'[f(x)]f'(x)$

The chain rule can continue on for multiple derivative hence the name of the rule.

Example 3.5-A
$h(x) = e^{sin(4x)}$
$h'(x) = e^{sin(4x)} cos(4x) 4$

The chain rule was applied twice in this example. The sine function was inside the exponential function and therefore its derivative was multiplied outside of the exponential function. The chain continues on because the trig function, now a cosine has a parameter that can be derived. The derivative of what is inside the trig function is multipled outside the trig function.

In a sense, every time you derive, the chain rule is being applied.

Example 3.5-B
$f(x)=x^2$

The function can be viewed as: $f(x) = g(x)^2$, where $g(x) = x$

$f'(x)=2g(x)x'$
$f'(x)=2x$

6. ## Lesson 3.6 - The Product Rule

Lesson 3.6 - The Product & Quotient Rules
What happens when you have a product of two terms that needs to be derived such as:

$f(x)=x^3\sin(x)$

The derivative is NOT:

$f(x)=3x^2\cos(x)$

The product needs to broken down for each of the two terms. The rule is as follows...

Given two functions as a product: $h(x)=f(x)*g(x)$.

The derivative of the product is the derivative of the first function multiplied with the second function PLUS the first function multiplied with the derivative of the second function. If that sounds confusing, the math should clear it up...

$h'(x)=f'(x)*g(x) + f(x)*g'(x)$

This shouldn't be difficult to remember. You take care of deriving the first term as a product with the other function. You then do this again, but switch the function you are deriving to the second one.

Example 3.6-A
$f(x)=x^3\sin(x)$

We can split the function into terms like so: $f(x)=x^3 * \sin(x)$

$f'(x)=3x^2 *\sin(x) + x^3 \cos(x)$

You are done! The product rule is as simple as that type of split. Still confused? Let's analyze the function a bit deeper...

$f(x)=x^3\sin(x)$

Let $g(x)=x^3$ and $h(x)=\sin(x)$, so $f=gh$

$f' = g'h + h'g$

$g'(x) = 3x^2$
$h'(x) = \cos(x)$

$f' = 3x^2 \sin(x) + \cos(x) x^3$

You may be asking yourself why the product rule exists as it does. In short, these notes do not dive into Analysis, but a short introduction may be worthwhile. The product rule, also known as Leibniz's law, can be shown using the rigorous proof via the definition of the derivative or through this alternate proof using logarithms.

Let $f=uv$ where $u, v > 0$, then:

$\ln|f| = \ln|u|+\ln|v|$

Differentiating both sides gives:

$\frac{f'}{f}=\frac{u'}{u}+\frac{v'}{v}$

Multiply both sides by $f$ which is $uv$

$f'=\frac{u'f}{u}+\frac{v'f}{v}$

$f'=\frac{u'uv}{u}+\frac{v'uv}{v}$

$f'=u'v+v'u$

Voila!

The Quotient Rule
Now, the Quotient Rule is merely an extension of the product rule since division is another form of multiplication. It is your choice whether or not you would like to remember an additional rule. If it is helpful for you, then by all means go ahead.

Say a function is of the form: $f=\frac{u}{v}$

It's derivative is: $f=\frac{u'v - v'u}{v^2}$

I'll skip the proof of the rule, but I will show you how the quotient rule is just a simplified version of the product rule.

$f=\frac{u}{v}$

$f=uv^{-1}$

Product of two terms, now follow the product rule like so..

$f'=u'v^{-1} - uv^{-2}v'$

This simplifies to the following...

$f'=\frac{u'}{v} - \frac{uv'}{v^2}$

$f'=\frac{u'v}{v^2} - \frac{uv'}{v^2}$

$f=\frac{u'v - v'u}{v^2}$

Example 3.6-B
$f(x)=\frac{x^2}{e^x}$

Let $u(x)=x^2, v(x)=e^x$, then $u'(x)=2x, v'(x)=e^x$

According to the quotient rule:

$f=\frac{u'v - v'u}{v^2}$

$f'(x) = \frac{2x e^x - e^x x^2}{e^{2x}}$

$f'(x) = \frac{x(2-x)}{e^x}$

7. ## Lesson 3.7 - Implicit Differentiation

Lesson 3.7 - Implicit Differentiation
Thus far, we have only derived one parameter of our equations, specifically the input, or as it is commonly used, the variable x. The derivative of y is simply $\frac{dy}{dx}$ also known as $y'$ or $f'(x)$. What if we are differentiating $y^2$ with respect to x? What do we do? We use what is known as implicit differentiation.

Implicit differentiation is simply another application of the chain rule.

The derivative of $y$ with respect to x is simply $y'$, but the derivative of $y^2$ w.r.t x is $2y*y'$. Y is derived just as x was, but by following the chain rule, we must tack on a $y'$.

Example 3.7-A
$y^2 + x^2 = 16$

$y^2 = 16-x^2$

$y=\pm\sqrt{16-x^2}$

$y'=\frac{\pm 1(2x)}{2\sqrt{16-x^2}}$

$y'=\frac{\pm x)}{\sqrt{16-x^2}}$

Now using implicit differentiation...

$y^2 = 16-x^2$

Derive each term with respect to x.

$2y*y' = 2x$

$y' = \frac{x}{y}$

Plug y back into the equation.

$y' = \frac{\pm x}{\sqrt{16-x^2}}$

Both methods join together for the same answer just like the product and quotient rules do as you learned in Lesson 3.6.

Example 3.7-B
Find the equation of the tangent line to:
$x^2+y^2=16$
at the point $(3, \sqrt{5})$.

We found the derivative in the last example, but it has a choice of signs. Our point is in the first quadrant, so y must be positive.

$y=\sqrt{16-x^2}$

$y' = \frac{x}{y}$

$y' = \frac{x}{\sqrt{16-x^2}}$

The slope of the tangent line is equal to the value of the derivative at the point $(3, \sqrt{5})$.

$y' = \frac{3}{\sqrt{16-3^2}}$

$m \Rightarrow y' = \frac{3}{\sqrt{5}}$

Using the slope-intercept form of a line, we have:

$y=\frac{3}{\sqrt{5}}x+b$

We need to plug the x-y coordinates $(3, \sqrt{5})$ to find the y-intercept.

$\sqrt{5}=\frac{3}{\sqrt{5}}3+b$

$b=\sqrt{5}-\frac{9}{\sqrt{5}}$

Our equation for the tangent line is finished, it is $y=\frac{3}{\sqrt{5}}x+\sqrt{5}-\frac{9}{\sqrt{5}}$ which simplifies to:

$y=\frac{3x-4}{\sqrt{5}}$

Implicit differentiation finds the most use when an equation has more than one term with the output variable, or y, in it.

Example 3.7-C
$y=x\sin(y)+\cos(x)$

Using implicit differentiation (and prior rules!) we get the following:

$y'= \sin(y) + x\cos(y)y'-\sin(x)$

Solving for y'

$y'(1-x\cos(y))= \sin(y) -\sin(x)$

$y'= \frac{\sin(y) -\sin(x)}{1-x\cos(y)}$

This is a differential equation. Plugging y back into it will do us no good and we will end up in an infinite loop of subbing $y=x\sin(y)+\cos(x)$ back into the equation.

Example 3.7-D
$y=x^x$

Remember I said we would go over this problem? Contrary to popular belief, this is not solved using the power rule: $y' \ne (x)x^{x-1}$. It is also not solved like you would derive an exponential: $y' \ne ln|x|x^x$

The trick is to use the properties of the log combined with implicit differentiation.

Take the log of both sides.

$\ln(y)=\ln(x^x)$

This brings down our exponent, x, and outside of the log as a coefficient.

$\ln(y)=x\ln(x)$

Now, take the derivative with respect to x.

$\frac{1}{y}y'=\ln(x)+\frac{x}{x}$

$y'=y(\ln(x)+1)$

Re-substitute our original equation of $y=x^x$ and we are done.

$y'=x^x(\ln(x)+1)$

8. Side note... I would like to have these notes posted on my website. Is it easier to install LaTeX on my site or upload images (of each equation) on to the server?

9. In addition to my Calculus I notes, Galactus has offered to post some notes & shortcuts on topics in Calculus II & Calculus III ranging from Integration by Parts, Partial Differentiation, Multiple Integrals and LaGrange Multipliers.

10. I am going to post a classic optimization problem. Solving it using calculus and withour using calculus. I was looking through my notes and came across this one.

Find the dimensions of the cone of minimum vloume that can be circumscribed about a sphere of radius R
First, my calculus method:

From the diagram, we let r=radius of the cone.

x+R=AO=height of cone. Where x is the distance from the apex of the cone to the center of the sphere.

From the similar triangles, we get:

$\frac{r}{R}=\frac{R+x}{\sqrt{x^{2}-R^{2}}}$

Solving for r, we get $r=\frac{R^{2}(x+R)}{x-R}$

Now, sub this into the formula for the volume of a cone, $V=\frac{\pi}{3}r^{2}h$:

$V=\frac{\pi}{3}\left(\frac{R^{2}(x+R)}{x-R}\right)^{2}(x+R)$

Differentiate wrt x:

$\frac{dV}{dx}=\frac{\pi}{3}\left(\frac{(x+r)(x-3r)r^{2}}{(x-r)^{2}}\right)$

Set to 0 and solve for x:

$(x+r)(x-3r)r^{2}=0$

Therefore, $x=-r, \;\ x=3r, \;\ x=0$

Our only viable solution is $x=3r$

Which tells us that the height is $h=3r+r=4r$

$r=\frac{R(R+3R)}{\sqrt{(3R)^{2}-R^{2}}}=\sqrt{2}R$

So, the minimum is achieved when the height is $4R$ and the radius is $\sqrt{2}R$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Now, the fun method. The one NOT using calculus. I worked this out some years back as a challenge problem. Took a while, I must admit.

Again, from the diagram:

height of cone=AC

h-2R=AE

radius of cone = CD = r

y = AF

From circle geometry, $AE\cdot{AC}=(AF)^{2}$

$(h-2R)h=y^{2}$

By similar triangles:

$\frac{y}{R}=\frac{h}{r}$

$yr=Rh$

$r=\frac{Rh}{y}$

Sub r into the cone volume formula as we done before:

$V=\frac{\pi}{3}\left(\frac{Rh}{y}\right)^{2}h$

$=\frac{{\pi}R^{2}}{3}\left(\frac{h^{2}}{h-2R}\right)$

Now, here is where we could differentiate. But....we don't want to do that.

Let's try and hammer it into shape with some algebra.

Looking at $\frac{h^{2}}{h-2R}$

$\frac{h^{2}}{h-2R}=\frac{1}{\frac{1}{h}-\frac{2R}{h^{2}}}$

Do some completing the square-type action:

$\frac{1}{\frac{1}{h}-\frac{2R}{h^{2}}}=\frac{1}{\frac{1}{8R}-\left(\frac{1}{2\sqrt{2R}}-\frac{\sqrt{2R}}{h}\right)^{2}}$

Now, the minimum is achieved when $\frac{1}{2\sqrt{2R}}-\frac{\sqrt{2R}}{h}=0$

Solving for h we see $h=4R$

Subbin this into r above we see $r=\sqrt{2}R$

Just as with the calculus method.

I hope this was of some help.

11. Galactus, I love the multiple methods approach!

12. Thanks, I will post some more as I get time. The related rates with the cone on its side instead of apex down was a good one I posted sometime back. That would be a good one to include in here.
Also, I have a good one involving the length of an amusement park ride we can do with polar or spherical coordinates.
I will try and post those soon.

Here is the link to the cone problem. I thought this was a goody.

http://www.mathhelpforum.com/math-he...e-problem.html

13. ## partial fraction using complex analysis

Here is a cool, easy-to-use, method of doing partial fractions by using a little

complex analysis. Don't be put off by the words 'complex and analysis'. It's

easy and a nice alternative to the standard method of using A and B and

what not.

Let's use the example $f(x)=\frac{x^{2}}{x^{2}-6x+8}$

When we factor the denominator as we would by the standard method, we get

$f(x)=\frac{x^{2}}{(x-4)(x-2)}$

Now, f(x) has simple poles at x=4 and x=2. See?. Those are the values that result in division by 0.

Now, we take the limits as such:

$\lim_{x\rightarrow{4}}(x-4)\cdot\frac{x^{2}}{(x-4)(x-2)} = \frac{x^{2}}{x-2} = \boxed{8}$

The other:

$\lim_{x\rightarrow{2}}(x-2)\cdot\frac{x^{2}}{(x-4)(x-2)} = \frac{x^{2}}{x-4} = \boxed{-2}$

There, we have the numerators in our partial fraction decomposition.

And we get:

$\boxed{r(x)=\frac{-2}{x-4}+\frac{8}{x-2}}$

Now, f(x)-r(x) has no poles in the complex plane, but the function is bounded so we must have a constant.

Since $\lim_{x\rightarrow{\infty}}[f(x)-r(x)]=1$

we have:

$f(x)=1+r(x)$

$\boxed{\frac{-2}{x-4}+\frac{8}{x-2}+1}$

I hope you find this helpful and a nice/fun alternative to doing PFD's.

14. Originally Posted by galactus
Here is a cool, easy-to-use, method of doing partial fractions by using a little

complex analysis. Don't be put off by the words 'complex and analysis'. It's

easy and a nice alternative to the standard method of using A and B and

what not.

Let's use the example $f(x)=\frac{x^{2}}{x^{2}-6x+8}$

When we factor the denominator as we would by the standard method, we get

$f(x)=\frac{x^{2}}{(x-4)(x-2)}$

Now, f(x) has simple poles at x=4 and x=2. See?. Those are the values that result in division by 0.

Now, we take the limits as such:

$\lim_{x\rightarrow{1}}(x-4)\cdot\frac{x^{2}}{(x-4)(x-2)} = \frac{x^{2}}{x-2} = \boxed{-1}$

The other:

$\lim_{x\rightarrow{2}}(x-2)\cdot\frac{x^{2}}{(x-4)(x-2)} = \frac{x^{2}}{x-4} = \boxed{-2}$

There, we have the numerators in our partial fraction decomposition.

And we get:

$\boxed{r(x)=\frac{-2}{x-1}+\frac{8}{x-2}}$

Now, f(x)-r(x) has no poles in the complex plane, but the function is bounded so we must have a constant.

Since $\lim_{x\rightarrow{\infty}}[f(x)-r(x)]=1$

we have:

$f(x)=1+r(x)$

$\frac{-2}{x-1}+\frac{8}{x-2}+1$

I hope you find this helpful and a nice/fun alternative to doing PFD's.
The only thing I dont understand is where the 1 came from in the first limit...why isnt it 4?

15. Galactus (Cody),

I have never seen this method for partial fractions. I have a couple of questions....

$
\lim_{x\rightarrow{1}}(x-4)\cdot\frac{x^{2}}{(x-4)(x-2)} = \frac{x^{2}}{x-2} = \boxed{-1}
$
Where did you get the limiting value of x being one?

$
\boxed{r(x)=\frac{-2}{x-1}+\frac{8}{x-2}}
$