Solução_Calculo_Stewart_6e

F.
92 ¤ CHAPTER 3 DIFFERENTIATION RULES
TX.10
79. y = x 2 ⇒ y 0 =2x, so the slope of a tangent line at the point (a, a 2 ) is y 0 =2a and the slope of a normal line is −1/(2a),
for a 6= 0. The slope of the normal line through the points (a, a 2 ) and (0,c) is a2 − c − c
a − 0 ,soa2 = − 1
a 2a
⇒
a 2 − c = − 1 ⇒ a 2 = c − 1 . The last equation has two solutions if c> 1 , one solution if c = 1 , and no solution if
2 2 2 2
c< 1 2 . Since the y-axis is normal to y = x2 regardless of the value of c (thisisthecasefora =0), we have three normal lines
if c> 1 2 and one normal line if c ≤ 1 2 .
3.2 The Product and Quotient Rules
1. Product Rule: y =(x 2 +1)(x 3 +1) ⇒
y 0 =(x 2 +1)(3x 2 )+(x 3 + 1)(2x) =3x 4 +3x 2 +2x 4 +2x =5x 4 +3x 2 +2x.
Multiplying first: y =(x 2 +1)(x 3 +1)=x 5 + x 3 + x 2 +1 ⇒ y 0 =5x 4 +3x 2 +2x (equivalent).
3. By the Product Rule, f(x) =(x 3 +2x)e x ⇒
f 0 (x)=(x 3 +2x)(e x ) 0 + e x (x 3 +2x) 0 =(x 3 +2x)e x + e x (3x 2 +2)
= e x [(x 3 +2x)+(3x 2 +2)]=e x (x 3 +3x 2 +2x +2)
x 2 d
5. By the Quotient Rule, y = ex
⇒ y 0 = dx (ex ) − e x d
dx (x2 )
= x2 (e x ) − e x (2x)
= xex (x − 2)
= ex (x − 2)
.
x 2 (x 2 ) 2 x 4 x 4 x 3
The notations
PR
⇒
and
QR
⇒
indicate the use of the Product and Quotient Rules, respectively.
7. g(x) = 3x − 1
2x +1
QR
⇒ g 0 (2x + 1)(3) − (3x − 1)(2) 6x +3− 6x +2 5
(x) = = =
(2x +1) 2 (2x +1) 2 (2x +1) 2
9. V (x) =(2x 3 +3)(x 4 − 2x)
PR
⇒
V 0 (x) =(2x 3 + 3)(4x 3 − 2) + (x 4 − 2x)(6x 2 )=(8x 6 +8x 3 − 6) + (6x 6 − 12x 3 )=14x 6 − 4x 3 − 6
1
11. F (y) =
y − 3
(y +5y 3 )= y −2 − 3y −4 y +5y 3 PR
⇒
2 y 4
F 0 (y) =(y −2 − 3y −4 )(1 + 15y 2 )+(y +5y 3 )(−2y −3 +12y −5 )
=(y −2 +15− 3y −4 − 45y −2 )+(−2y −2 +12y −4 − 10 + 60y −2 )
=5+14y −2 +9y −4 or 5+14/y 2 +9/y 4
13. y = x3
1 − x 2 QR
⇒
y 0 = (1 − x2 )(3x 2 ) − x 3 (−2x)
= x2 (3 − 3x 2 +2x 2 )
= x2 (3 − x 2 )
(1 − x 2 ) 2 (1 − x 2 ) 2 (1 − x 2 ) 2