Derivatives Avi Kulkarni, April 28 - May 02

For derivatives:
There are two types of notations:
d/dx f(x) or f '(x)
The d/dx notation is called Leibniz notation, created by Gottfried Wilhelm Leibniz , while the f '(x) notation was developed by Joseph Louis Lagrange
Useful derivatives
e^x d/dx = e^x
ln(x) d/dx = 1/x
b^x d/dx = ln(b) * b^x
logb(x) d/dx = 1/( x*ln(b) )
x^n d/dx = nx^(n-1)
sin(x) d/dx = cos(x)
cos(x) d/dx = -sin(x)
tan(x) d/dx = sec(x)^2
Useful rules for deriving
[f(x) + g(x)] d/dx = f'(x) + g'(x)
f(x) * g(x) d/dx = f'(x) * g(x) + f(x) * g'(x)
f(x) / g(x) d/dx = [g(x) * f'(x) - g'(x) * f(x)] / (g(x)^2)
f(x) ^ g(x) d/dx = [f(x) ^ g(x)] * [g'(x) * ln(f(x)) + g(x) * f'(x) / f(x) ]
f(g(x)) d/dx = f'(g(x)) * g'(x)
f'(x) = lim[h --> 0] [f(x + h) - f(x)] / h
red light means stop, green means go.

Derivatives [Avi Kulkarni, April 28 - May 02]

welcome to the wonderful world of differential calculus! This section of the notes is dedicated to building up you understanding of calculus and then utterly crushing your self esteem with brutal questions. have fun!
useful things to note before proceeding
-everything above
-continuity of functions
ok, lets get on with it.
Approximating the slope of the tangent of a function at some point by first principles means finding the slope of ever more approximate secants until x1 and x2 are REALLY close together. The slope of the tangent of a function at that same point can usually be found from first principles.
The derivative of a function f'(x) is a function that describes the slope of the tangent of f(x) for all values of x. to find the derivative from first principles some tricky algebra is usually required.
ex. a) if f(x) = x^3, find f'(x)
f'(x) = lim[h->0] ((x+h)^3 - x^3)/h
= lim[h->0] (x^3 + 3x^2h + 3xh^2 + h^3 - x^3)/h
= lim[h->0] (3x^2 + 3xh + h^2)
= 3x^2
b) find the slope of the tangent line at x=0
f'(x) = 3x^2
f'(0) = 3(0)^2
f'(0) = 0
There are rules derived from the first principle definition. for example, the power rule. These are very very convenient.
ex. if f(x) = 131071*x^233, find f'(x)
if you try this using first principles, it will take you a very long time. so just use the power rule
f'(x) = (233*131071)*x^232
= 30529543 * x^232
bonus points to anyone who wants to try and factor that last coefficient.
the fun part of derivatives is when you start mixing elementry functions. using the rules noted above, these can be broken down into derivatives you can easily find.
ex. f(x) = sin(2x+1) * 2^x, find f'(x)
let g(x) = sin 2x+1, and h(x) = 2^x, then f(x) = g(x) * h(x)
by product rule
f'(x) = g'(x) * h(x) + h'(x) * g(x)
f'(x) = sin'(2x+1) * 2^x + [ln 2 * 2^x] * sin(2x+1)
if we now let L(x) = sinx, and K(x) = 2x+1, then g(x) = L(K(x))
by chain rule
g'(x) = L'(K(x)) * K'(x)
g'(x) = 2 * cos (2x + 1)
substituting into the main equation
f'(x) = [2*cos(2x+1)] * 2^x + [ln 2 * 2^x]*sin(2x+1)
f'(x) = [2^x] * [ 2*cos(2x+1) + ln 2 * sin(2x+1)]
ok, now for you to try some
1) find the derivatives of the following functions
a) f(x) = 2 * x^57
b) f(x) = 1
c) f(x) = (sin x)^2 + (cos x)^2
d) f(x) = sinx / (2cosx)
e) f(x) = e^x + 123671238907891249467899807895890789897895689528794568943678344597805238905230798702379045082537803648968702348968790352.7
f) f(x) = e^(2^x)
g) f(x) = 5*(sinx/x^2 - 1)^3
h) f(x) = x^3 + 3x^2 + 3x + 1
i) f(x) = (x+1)^3
2) A focal point of a parabola is the point where any ray emmited from this point will reflect off of the parabola onto a path parallel to the y-axis. Lines reflecting from a point on a curve reflect off of the line tanget to the curve at that point.
Given the parabola y = x^2
a) find the focal point
b) Prove that all rays that are parallel to the y-axis reflect so that either they converge at the focus or appear to be emmited from the focus
Maxima and minima
One of the most useful properties of the derivative fuinction is that it lets you find when the slope of the tangent of a function is 0. this is useful for finding the local maxima or minima of a function. There is of course the possibility that a point of inflection is described instead of a maximum or minimum, but in this section we really don't care because the first order derivative is far more fun to play with.
ex. Given the function of the height of a ball is -5x^2 + 5x + 1, find the maximum height of the ball.
f(x) = -5x^2 + 5x + 1
f'(x) = -10x + 5
we know the maximum of the original function occurs when f'(x) is 0, so
0 = -10x + 5
x = 1/2
substituting in the original function
f(1/2) = -5(1/4) + 5(1/2) + 1
f(1/2) = 9/4
the maximum height.
The local maximum or minimum is not always the highest point or lowest point of a function, but it is the highest/lowest point in a given section.
ex. Find the maximum/minimum value of f(x) = (x+2)(x)(x-2)
f(x) = (x^2-4)x
f'(x) = (x^2 - 4)*1 + (2x) * x product rule
f'(x) = 3x^2 - 4
so when f'(x) = 0, x = ?
0 = 3x^2 -4
x = -2/sqrt(3), 2/sqrt(3)
back in the original expression, we then find that the maximum and minimum values are...
f(-2/sqrt(3)) = 16/(3*sqrt(3)) <----max
f(2/sqrt(3)) = -16/(3*sqrt(3)) <----min
everyone agree with me?
are you sure?
try f(4). see what happens.
f(4) = 6*4*2 = 48.
but f(4) > f(-2/root(3)). f(-2/root(3)) isn't a maximum value. nothing is, oh how trick questions are amusing. If I were to ask you for the LOCAL maxima/minima of f(x) then you would do that whole derivative thing. but I just asked for a maximum or minimum, so the answer you should give in this case is +-infinity.
What are local maxs/mins good for then? Well, we can use these values to find the relevent maximum and minimum for optimization problems.
ex. A box is to be made from a 12 x 12 cardboard sheet. The box is made by cutting four identical squares from the corners and then folding up the sides. find the maximum area of the box
so.
if x is the length of the cut off square.
Volume = lwh
h = x
w = 12 - 2x
l = 12 - 2x
V = 2*2*(6-x)(6-x)x
= 4*(6-x)(6x - x^2)
V d/dx = 4*-(6x - x^2) + 4*(6-2x)(6-x) by product rule
V d/dx = 4*(6-2x)(6-x) - 4x(6-x)
V d/dx = 4*(6-x)*(6 - 2x - x)
V d/dx = 12*(6-x)(2-x)
if V d/dx = 0, then
x = 6, 2
so
Vmax = 0, 128
obviously 128 is bigger, so that must be the maximum volume. Note that though x = 10 wil give a larger volume, this does not apply to the problem as the maximum allowed value for x is 6 (otherwise you will be cutting cardboard which no longer exists).
now for another evil question set, MWHAHAHAHAHA!!!!!!! (I don't get to do that ofter enough)
1. determine the maximum or minimum value for the following functions
a) f(x) = x^2 - 9
b) f(x) = x^3
c) f(x) = e^x
d) f(x) = 1/x
2. determine the local maxima/minima for the following functions
a) f(x) = (x+2)(x-2)(x-1)
b) f(x) = 1 + sqrt(1 + sqrt(1 + sqrt(1 + sqrt(1 + sqrt(1 + sqrt(1 + sqrt(1 + sqrt(1 + sqrt(1 + sqrt(1 + ...
c) f(x) = -5x^2 + 2x + 1
d) f(x) = 2sin(x)
e) f(x) = cos(x) - x
3. A triangle is defined by the x and y axes and a line that passes through the point (2,1). Determine the minimum area of the triangle, full solution required.