Page 103 - Calculus Workbook For Dummies

87
Chapter 6: Rules, Rules, Rules: The Differentiation Handbook

2
*x d tan e = 6 xe x 2 tan e x 2 sec e x 2
x
2
2
3
dx
Holy quadrupely nested quadruple nestedness, Batman! This is one for the Riddler.
d x 2 3
= ` tane j
dx
2 x l d
2
2
3
x
2
= 3 ` tane j $ ` tane j c because stuff = 3 stuff $ stufflm
dx
x l
2
2
2
2
2
x
= 3 tan e x 2 sec e $ ` e j c because d tan glob = sec glob $ i globlm
i
_
_
dx
2 l
2
2
2
x
2
x
= 3 tan e x 2 sec e $ e $ _ x i c because d e lump = e lump $ lumplm
dx
2
2
2
x
2
x
= 3 tan e x 2 sec e $ e $ 2 x
2
2
= 6 xe x 2 tan e x 2 sec e x 2
BAM!!!
POW!!
CRUNCH!
2
3
x
y If y - x = + y , y = l 2 x + 1
3 y - 1
2
1. Take the derivative of all four terms, using the chain rule (sort of) for all terms containing a y.
2
1
3 y y - l 2 x = + yl
2. Move all terms containing y’ to the left, all other terms to the right, and factor out y’.
2
1
yl
3 y y - = + 2 x
l
2
l
1 = +
y 3 _ y - i 1 2 x
3. Divide and voilà!
2 x + 1
y = l
y 3 2 - 1
4 ye x
x
A If y 3 + lny = 4 e , y = l 3 y + 1
Follow steps for problem 25.
1 x
y 3 + l y y = l 4 e
1 x
y 3 + y m = 4 e
l c
4 e x
y = l 1
3 + y
4 ye x
=
y 3 + 1