Page 104 - Calculus Workbook For Dummies

88 Part III: Differentiation

3
dy y - 2 xy + 1
3
2
x
C For x y = y x + y 5 + , find dx by implicit differentiation. y = l - 3 y x + x - 5
2
2
This time you’ve got two products to deal with, so use the product rule for the two products and
the regular rules for the other two terms.
2 l 2 3 l 3
_ x i y + x y = l _ y i x + y x + l y 5 + l 1
2
2
l
3
2 xy + x y = l 3 y y x + y + y 5 + l 1
2
2
3
l
1
x y - l 3 y y x - y 5 = l y + - 2 xy
2
2
3
y x - 3 y x - i y - 2 xy + 1
5 =
l _
3
y - 2 xy + 1
y = l
2
2
- 3 y x + x - 5
*C If y + cos y = sin x , find the slope of the curve at d π , 0n. The slope is zero.
2
2
3
5
10
You need a slope, so you need the derivative.
3 l
2
3
3
yl + 2 cosy $ _ - siny _ i y i = cos 5 x ^ i 10 xh
_
\ 1444 2444 3
Implicit 1444444 2444444 3
Chain Rule Chain Rule
Differentiation
] twice nested g
2
3
3
5
y + l 2 cosy - siny _i 3 y y = 10 x cos x 2
_
li
\
Implicit
Differentiation
3
y 1 - y 6 2 cosy 3 siny i = 10 x cos x 2
5
l _
10 x cos x 2
5
y = l 2 3 3
1 - y 6 cosy siny
You need the slope at x = π , y = 0, so plug those numbers in to the derivative. Actually, you
10
can save yourself a lot of work if you notice that the numerator will equal zero (because
2
π
cos 5 e o = 0) and the denominator will equal 1 (because y = 0). And thus the slope of the
10
curve at this point is zero. A tangent line with a zero slope is horizontal, and because this
tangent line touches the curve where y = 0, the tangent line is the x-axis.
4
D For y = x , find the 1st through 6th derivatives.
y = l 4 x 3
y = ll 12 x 2
lll
y = 24 x
( )
4
y = 24
( )
5
y = 0
( )
6
y = 0
Extra credit: y (2005 ) = 0
3
5
E For y = x + 10 x find the 1st, 2nd, 3rd, and 4th derivatives.
4
y = l 5 x + 30 x 2
3
y = ll 20 x + 60 x
lll
2
y = 60 x + 60
( ) 4
y = 120 x