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Class Notes - Calculus and Matrix Algebra for Business | 22M 017, Study notes of Mathematics

Material Type: Notes; Class: 22M - Calculus and Matrix Algebra for Business; Subject: Mathematics; University: University of Iowa; Term: Fall 2006;

Typology: Study notes

Pre 2010

Uploaded on 03/19/2009

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22M:017, Fall 2006
Lecture 13 (9/20/06)
Juan A. Gatica
We have seen the Chain Rule. We will spend a great deal of time today doing
examples, since this is an important topic.
Example. The limit
lim
h0
2h313
h
is of the form fx0.Find f,x0and the limit.
Solution.
Recall that
fx0lim
h0
fx0hfx0
h.
Thus we must have
fxx31
x02.
Then
fx1
2x31
3x2(using the Chain Rule).
Then
lim
h0
2h313
hf2
12
6
2.
Example. Do the same as in the previous example for
lim
h0
e1h21he2
h.
With the same reasoning:
fxex2x,x01.
Then
pf3
pf4

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22M:017, Fall 2006 Lecture 13 (9/20/06)

Juan A. Gatica

We have seen the Chain Rule. We will spend a great deal of time today doing examples, since this is an important topic.

Example. The limit

lim h 0

 2  h^3  1  3 h is of the form fx 0 . Find f, x 0 and the limit. Solution. Recall that

fx 0   lim h 0 fx^0 ^ h^ ^ fx^0  h

Thus we must have fx  x^3  1 x 0  2. Then fx  1 2 x^3  1

3 x^2 (using the Chain Rule).

Then

lim h 0

 2  h^3  1  3 h  f 2 

 (^126)  2. Example. Do the same as in the previous example for lim h 0 e  1 h^2  1 h (^)  e 2 h

With the same reasoning: fx  e x^2 x, x 0  1. Then

fx   2 x  1 e x^2 x f 1   3 e^2. Example. As above, find

lim h 0 log 3  3  h^2  3  h  log 3  12  h

Here fx  log 3 x^2  x, x 0  3. To find the derivative of log 3 x^2  x we MUST change the base in order to use the natural logarithm. We also must use the chain rule.

fx  log 3 x^3  x  1 ln 3  lnx^3  x

Then: fx  3 x

ln 3 x^3  x f 3   28 30 ln 3 

Thus:

lim h 0 log 3  3  h^2  3  h  log 3  12  h ^

30 ln 3 

Example. Find the derivative of the function fx  e x^  lnx^2  1 ^10. In this example, if we let hx  e x^  lnx^2  1 , gx  x^10 then fx  g  hx, so we can use the chain rule. Since hx  e x^  (^) x^22 x 1 , gx  10 x^9 we get:

fx  ghxhx  10 e x^  lnx^2  1 ^9 e x^  2 x x^2  1

In general if fx is a differentiable function and r is a real number and if gx  fxr^ is defined, then g is differentiable and:

gx  rfxr^1 fx.

Example. Find the slope of the line tangent to the graph of fx  x^2  x  1  23 at the point 1, 3 (^23) . The slope is f 1  and fx  23 x^2  x  1 ^ (^13)  2 x  1  so:

fx  e^

x (^)  exe x (^)  ex  e x (^)  exe x (^)  ex e x^  ex^2

 e^

x (^)  ex (^2)  e x (^)  ex 2 e x^  ex^2

 e^

x (^)  ex (^)  e x (^)  exe x (^)  ex (^)  e x (^)  ex e x^  ex^2

  2 e x 2 ex e x^  ex^2  4 e x^  ex^2

Example. Find the derivative of the function fx  x^2. Here we have to use the chain rule and change the base of the exponential:

fx  eln x^2 fx  2 x ln eln x^2  2 x ln x^2.