Problem Set 7
Due 2:00PM Sep. 12
1
Additional Differentiation Rules
We did not prove the following propositions yet. We are going to prove on Monday. Feel free to
use the following results for this problem set if needed.
Proposition 1.1 Suppose that f (x) = lnx. Then f 0 (x) = x1 .
Proposition 1.2 For any positive base b,
d x
dx a
= (ln a) (ax ).
Proposition 1.3
d
(sinx) = cosx
dx
d
(cosx) = −sinx
dx
d
(tanx) = sec2 x
dx
2
d
(cscx) = −cscx · cotx
dx
d
(secx) = secx · tanx
dx
d
(cotx) = −csc2 x
dx
Questions
1. Find the derivatives of the following functions.
• f (x) = e5x−4
Using the chain rule,
f 0 (x) = 5e5x−4
• f (x) = ln
√
3x2 + 2x
Using the chain rule,
1
1
][ (3x2 + 2x)−1/2 ][6x + 2]
+ 2x 2
1
1
1
= [√
]( )[ √
](2(3x + 1))
2
2
3x + 2x 2
3x + 2x
3x + 1
= 2
3x + 2x
f 0 (x) = [ √
3x2
1
• f (x) = e3x−2
Use the chain rule:
f 0 (x) = 3e3x−2
• f (x) =
(x−x2 )
e2x
Use the quotient, and the chain rule to fin
d
2x
dx (e ):
(1 − 2x)(e2x ) − (x − x2 )[e2x (2)]
(e2x )2
(e2x )[(1 − 2x) − 2(x − x2 )]
=
(e2x )(e2x )
1 − 4x + 2x2
=
.
e2x
f 0 (x) =
• f (x) = ln(g(x2 ))
Use the chain rule, since g is a function:
f 0 (x) = [
2x[g 0 (x2 )]
1
0 2
][g
(x
)](2x)
=
g(x2 )
g(x2 )
• f (x) = eg(2x)
Use the chain rule, since g is a function:
f 0 (x) = [eg(2x) ][g 0 (2x)][2]
• f (x) = g(x) + h(y)
Since f is a function of x only, then h(y) is some constant.
f 0 (x) = g 0 (x)
• f (x) = (xδ + aδ )γ
The only variable is x and a, δ, γ are constants. Use the chain rule:
f 0 (x) = γ(xδ + aδ )γ−1 (δxδ−1 )
= δγxδ−1 (xδ + aδ )γ−1
2
2. Find the derivatives of the following functions.
• f (x) = 186.5 , Answer: 0
• f (x) = 5x − 1 , Answer: 5
• f (x) = x2 + 3x − 4 , Answer: 2x + 3
• f (t) = 14 (t4 + 8) , Answer: t63
• f (x) = x−2/5 , Answer: − 52 x−7/5
• V (r) = 43 πr3 , Answer: 4πr2
• Y (t) = 6t−9 , Answer: −54t−10
√
√
• G(x) = x − 2ex , Answer: 1/(2 x) − 2ex
• F (x) = ( 12 x)5 , Answer: 5/32x4
1
x2
• g(x) = x2 +
• y=
, Answer: 2x − (2/x3 )
√
√
√
, Answer: 32 x + (2/ x) − 3/(2x x)
x2 +4x+3
√
x
• y = 4π62 , Answer: 0
• y = ax2 + bx + c , Answer: 2ax + b
√
4 3
1
• v = t2 − √
t )
4 3 , Answer: 2t + 3/(4t
t
3. Differentiate
• f (x) = x2 ex , Answer: x(x + 2)ex
• y=
ex
x2
, Answer: (x − 2)ex /x3
3x−1
2x+1
= (2x3
• g(x) =
• V (x)
• F (y) =
• y=
• y=
1
y2
, Answer: 5/(2x + 1)2
+ 3)(x4 − 2x) , Answer: 14x6 − 4x3 − 6
− y34 (y + 5y 3 ) , Answer: 5 + 14/y 2 + 9/y 4
t2
, Answer: 2t(1 − t)/(3t2
3t2 −2t+1
(r2 − 2r)er , Answer: (r2 − 2)er
√
v 3 −2v v
v
y = x4 +x1 2 +1
f (x) = x+x c
x
− 2t + 1)2
• y=
√
, Answer: 2v − 1/ v
•
, Answer: −(4x3 + 2x)/(x4 + x2 + 1)2
•
, Answer: 2cx/(x2 + c)2
4. Differentiate
• y = sin4x , Answer: 4cos4x
• y = (1 − x2 )1 0 , Answer: −20x(1 − x2 )9
√
√
√
• y = e x , Answer: e x /(2 x)
• F (x) = (x3 + 4x)7 , Answer: 7(x3 + 4x)6 (3x2 + 4)
√
2+3x2
• F (x) = 4 1 + 2x + x3 , Answer: 4(1+2x+x
3 )3/4
• g(t) =
1
(t4 +1)3
3
, Answer: − (t412t
+1)4
• y = cos(a3 + x3 ) , Answer: −3x2 sin(a3 + x3 )
3
• y = e−mx , Answer: −me−mx
• g(x) = (1 + 4x)5 (3 + x − x2 )8 , Answer: 4(1 + 4x)4 (3 + x − x2 )7 (17 + 9x − 21x2 ))
• y = (2x − 5)4 (8x2 − 5)−3 , Answer: 8(2x − 5)3 (8x2 − 5)−4 (−4x2 + 30x − 5)
• y = xe−x
2
2
, Answer: e−x (1 − 2x2 )
• y = excosx , Answer: (cosx − xsinx)excosx
q
1/2 (z + 1)3/2 ]
• F (z) = z−1
z+1 , Answer: 1/[(z − 1)
• y=
√ r
r2 +1
, Answer: (r2 + 1)−3/2
• y = tan(cosx) , Answer: −sinxsec2 (cosx)
• y = 2sinπx , Answer: 2sinπx (πln2)cosπx
• y = (1 + cos2 x)6 , Answer: −12cosxsinx(1 + cos2 x)5
• y = sec2 x + tan2 x , Answer: 4sec2 xtanx
• y = cot2 (sinθ) , Answer: −2cosθcot(sinθ)csc2 (sinθ)
√
p
√
√ √x)
• y = x + x , Answer: 1+1/(2
2 x+ x
√
√
√
√
• sin(tan sinx) , Answer: cos(tan sinx(sec2 sinx)[1/(2/ sinx)](cosx)
4