# Calculus Extra Credit Assignment

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MATH EXTRA CREDIT PAPER

This purpose of this paper is to give some examples of derivatives. These problems also show the different types of ways to solve for derivatives including the power rule, the quotient rule, product rule, and chain rule.

PROBLEM 1:

Y= 2xn +x3-n+13

2xn-1 + (3-n)x2-n

PROBLEM 2:

Y= (x-1)(x2+3)

FOIL: x3-x2+3x-3

Then take the derivative and apply the power rule:

3x2-2x+3

PROBLEM 3:

Y= x4/4 - x-8

Then take the derivative and apply the power rule:

X3+ 8x-9

PROBLEM 4:

Y= x/2 + 2/x

Apply the power rule, and take the x in the denominator

and put it on top so that portion of the equation equals:

-2/x2

1/2- 2/x2

PROBLEM 5:

Y = (x+2) 2-5x3

Foil out (x+2) 2

So that it equals: x2+4x+4

Simplify (and include -5x3)

-15x2+2x+4

PROBLEM 6:

Y= x(x2-3)(x3+1)

First foil out: (x2-3)(x3+1)

So that it equals:

x5-3x3+x2-3

then multiply x throughout

6x5-12x3+3x2-3

PROBLEM 7:

Y= (ax+1) 2 – 2ax3

Where a= const

First foil out:

(ax+1) (ax+1) and you get:

ax2 + ax + ax + 1 -2ax3

then simplify:

-2ax3+ax2+2ax+1

then apply the power rule and you should get:

-6ax2+2a2x+2a

PROBLEM 8:

Y= (3x+1) 2

Use chain rule to achieve the derivative,

The outer layer is the square ad the inner layer is (3x+1). First you must differentiate “the square” first, and leave (3x+1). After that, you should differentiate (3x+1):

D (3x+1) 2

= 2(3x+1) 2-1 D(3x+1)

= 2(3x+1) (3)

=6(3x+1)

=18x+6

PROBLEM 9:

D((x2 +2x+1)(x3 +5))

(2x+2)(x3 +5) + (x2 +2x+1)(3x2)

2x4+10x+2x3+10+3x4+6x3+3x2

5x4+8x3+3x2+10x+10

PROBLEM 10:

D (x2+1)/(3x)

= (3x(2x)-(x2+1)(3))/ ((3x) 2)

(6x2-3x2-3)/ (9x2)

(3x2-3)/ (9x2)

PROBLEM 11:

F(x)= (x2+1)(2x3-4)

(x2+1)(6x2) + (2x)(2x3-4)

6x4+6x2+4x4-8x

10x4+6x2-8x

PROBLEM 12:

F(x)= (x2+1)/(2x2-1)

= ((2x2-1)(2x)- (x2+1)(4x))/ ((2x2-1) 2)...

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