AMA106, Semester 1, 2009/2010, solutions for Problem set 2.
1 1. (a) f (x) = x 2 (2 − 7x2 ) √ 2−14x2 2−7x2
−1 2
(−14x) +
√
2 − 7x2 = −7x2 (2 − 7x2 )
1 4− 2√x √ 2 4x− x
−1 2
+
√
2 − 7x2 =
−7x2 +2−7x2 √ 2−7x2
=
(b) f (x) = 1 (4x − 2
√
x)
−1 2
(4 − 1 x 2 ) = √ 2
1
(c) f (x) = 3(− sin(πx2 ))2xπ = −6xπ sin(πx2 ) (d) f (x) = − sin(sin(x)) cos x (e) f (x) = (f) f (x) = (g) f (x) = (h) f (x) = (i) f (x) = (j) f (x) = (k) Let y = (l) f (x) = (m) f (x) = (n)
1 x(5−2x) 2
3
−
1 x2 (5−2x) 2
1
=
x−5+2x x2 (5−2x) 2
3
=
3x−5 x2 (5−2x) 2
3
2 4 −4x2 +1) (x4 +1)4 2(x2 +1)2x−(x2 +1)2 4(x4 +1)3 4x3 = 4x(x +1)(−3x 5 (x4 +1)8 (x4 +1) 1 1 1 1 1 cos( x ) + x(− sin( x ))( −1 ) = cos x + x sin x x2 e−x cos(ax)a + sin(ax)(−e−x ) = e−x (a cos(ax) − sin(ax))
2x x2 +1 1 x ln x 35x−7
1 dy ⇒ ln y = (5x − 7) ln 3 ⇒ y ( dx ) = 5 ln 3 ⇒
dy dx
= 5y ln 3 = 5 ln 3(35x−7 )
(1+6x )6−6x(6x ln 6) (1+6x )2
=
6+6x+1 −6x+1 x ln 6 (1+6x )2
6x(x2 +1)2 (2x+1)5 (3x−2)3 −(x2 +1)3 [10(2x+1)4 (3x−2)3 +9(2x+1)5 (3x−2)2 ] (2x+1)10 (3x−2)6 4x+5 4x+5 Simplifying the function first, f (x) = 2+ 1 3 = 2+ 13x = 2(4x+5)+3x = 11x+10 5 4x+5
4+ x
Differentiating, f (x) = 2. (a) By chain rule, (y◦u)(x) = (b)
dy du 1+( dy dx 1
4(11x+10)−11(4x+5) (11x+10)2
=
−15 (11x+10)2
=
dy du
·
1 )2 1+x2
=
4x(1+x2 ) dy du dy −2u −2x du dx , du = (1+u2 )2 , dx = (1+x2 )2 , ⇒ dx = [(1+x2 )2 +1]2 2 2 2 )−(1+x2 2 2 (1+x2 )2 , d(y◦u)(x) = [(1+x ) +1]2x2(1+x2 )2 +1]2 ) 2(1+x )2x dx (1+x2 )2 +1 [(1+x
=
4x(1+x2 ) [(1+x2 )2 +1]
=
(y ◦
dy −2 1 −1 , du = 2√x+1 , ⇒ dx = √x+1(√x+1−1)2 (u−1)2 dx √ −1 u)(x) = √x+1+1 , ⇒ d(y◦u)(x) = √x+1(√x+1−1)2 dx x+1−1
3. (a) f (x) = 3Ax2 + 2Bx + C. The curve intersects the point (1,0), so 0 = A + B + C + D. Furthermore, the slope of the tangent line at (1,0) is 3, so f (1) = 3, hence 3 = 3A + 2B + C . The curve also intersects (2,9) where the slope of the tangent line is 18....