When an equation is differentiated we are finding an equation which
shall enable us to calculate the rate of change on the graph of that
equation at any point we may chose. In many situations it is important
to know the rate of change at a chosen point on a graph of a function.
for a a graph of a function which represents a staight line, it is
quite simple to identify the grident of that line either
geomettrically or algebralically. If a graph of a function is a curve,
the graident is different at different points on the curve. It is
therfore much more difficult to calculate the rate of change at any
point on the curve. This is why we use the rules of differentiation to
minipulate a given function and therfore enable ourselves to
calculatethe rate of change at any point.

1/ examples 

differentiate                                         

 y=(2x+3)^2

let u = 2x+3

then du/dx =2 and y=u^2

then dy/du = 2u

since dy/dx = dy/du * du/dx

by substituting dy/dx

= 2u *2 = 4(2x+3)

2/

y= (x^2 -3x+5)^3

let u = x^2 - 3x+5

du/dx = 2x-3

y = u^3

dy/du = 3u^2

again substituting

dy/dx = dy/du * du/dx

= 3u^2 8 (2x-3)

= 3(x^2 - 3x +5)^2 (2x - 3)

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answer 4 questions show all working and simplify as much as possible
giving reasons for answers.

differentiate the following 4 questions 

1/ y = (x^2 +2)(3x - 1)

2/ y = (2x - 4)^2

3/ i = 20 sin (100 pi t +pi/3)

4/ i = 50 cos (50 pi t - pi/6)

Thank you for the questions. Here are your answers.

Question 1
----------
We differentiate 

 y= (x^2+2)(3x-1)

with respect to x .

We will expand the function into a polynomial and the differentiate
that polynomial:

y=(x^2+2)(3x-1)
 = (x^2+2)3x + (x^2+2)(-1)
 = 3x^3+6x -x^2 -2
 = 3x^3 -x^2 +6x -2

Now can just differentiate this polynomial directly:

dy/dx = 9x^2 - 2x + 6

This is the answer.

Question 2
---------

In this question and the succeeding questions, we have to find an
appropriate u so that we can use the chain rule to simplify the
computation.

We differentiate

y= (2x-4)^2

with respect to x .

We set u=2x-4 .

Then y=u^2 .

Hence, 

 dy/du = d(u^2)/du = 2u

and 

  du/dx = d(2x-4)/dx = 2 .

Hence, by the chain rule:

dy/dx = (dy/du) (du/dx)
      = 2u (2) 
      = 4u
      = 4(2x-4)
      = 8x-16


Question 3
----------
We differentiate 

i=20 sin (100 pi t + pi/3) 

with respect to t.

We let u= 100 pi t + pi/3 .

Then du/dt = 100 pi .

We have 

i=20 sin (u)

so 

di/du = 20 cos (u) .

Hence by the chain rule,

di/dt = (di/du) (du/dt)
      = 20 cos(u) 100 pi
      = 2000 cos(u)
      = 2000 cos (100 pi t + pi/3)

Question 4
----------

We differentiate

i = 50 cos (50 pi t - pi/6)

with respect to t.

We set

u= 50 pi t - pi/6 .

Then

du/dt = 50 pi

We have

i = 50 cos (u)

so

di/du = -50 sin (u)

Hence by the chain rule:

di/dt= (di/du)(du/dt)
     = -50 sin (u) 50 pi
     = -2500 sin (u)
     = -2500 sin (50 pi t - pi/6)


======================
Plots:

I have made simple plots of  each function and its derivative. This
might help you to visualize what is going on with the functions and
the derivatives.

One simple thing to observe in these plots is that whenever the
function has a local maximum or minimum (that is, at the top of a
"hill" or the bottom of a "valley" in the graph of the function) the
value of the deriative is always 0 there. This is a general property
of the deriviative - it's one of the important uses of deriviatives,
actually, as it is used in physics in a lot.

The plots are at:

http://www.rbnn.com/google/differentiation/1.jpg
http://www.rbnn.com/google/differentiation/2.jpg
http://www.rbnn.com/google/differentiation/3.jpg
http://www.rbnn.com/google/differentiation/4.jpg

for questions 1 through 4 respectively. I used Matlab to generate
these plots. Matlab is an excellent program, see
http://www.mathworks.com , and I highly recommend it for playing
around with mathematical functions.

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