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MTH124-005 SS17 Derivative Worksheet
Name:
The purpose of this worksheet is to provide an opportunity to practice differentiation
formulas for section 005. It will not be graded and you are not expected to finish in class.
There are commonly used formulas after the problems, some of these problems might be
challenging, if you have questions, feel free to ask me after class, or come to my office during
office hours.
Exercises
2 2 x
(1) f(x) = x +2x+1 (13) h(x) = (x +x+1)(4 )
xlnx
(2) g(x) = 3xex 2
(14) t(x) = ln(x2 + 3x)ex −x
(3) h(x) = ln(x2 + x) (15) n(x) = 1
lnx+x
(4) t(x) = 3x3e7
(16) a(t) = t(t3 + t + et)
x+1
(5) n(x) = x−1 (17) f(u) = e7+ln2+1 −1
u
2
t √
(6) a(t) = te (18) g(x) = x3 +2x+1
2
(7) f(u) = u 5y 1
u
ln(1+e ) (19) h(y) = e +ln(2y)+ y5
√
4 4 2
(8) g(x) = e 3x +3x +1 (20) s(t) = t3+7t+1
t
(9) h(y) = 1 x 2
2
(7y) (21) f(x) = ln(e (x +1))
3 2 2 7 5
(10) s(x) = (5x +2x +2)ln(x) (22) g(x) = e(x +3) (x−1)
3x
e +x
2 100 (23) h(x) = lnx−ln(1)
(11) f(x) = (x +x) x
(12) g(x) = (3x2 +x+1)exlnx x √
(24) t(x) = y(2 − 5)
2
(25) n(x) = (x+1)4(x−1)4x3 q 2
(28) g(x) = x +x
lnx+1
(26) a(t) = t2−1
t2 2
e −1 (29) h(y) = |y |
4
(27) f(u) = u 1
u (30) s(t) =
ln(1+e ) |t|
formulas
(1) d c = 0
dx
(2) d xn = nxn−1
dx
(3) d (mx+b) = m
dx
(4) d ex = ex
dx
d kx kx
(5) dxe =ke
d x x
(6) dxa = a ln(a)
(7) d ln(x) = 1
dx x
(8) d [cf(x)] = cf′(x)
dx
(9) d [f(x) ±g(x)] = f′(x)±g′(x)
dx
(10) d [f(x)g(x)] = f′(x)g(x) + f(x)g′(x) Product Rule
dx
d ′ ′ ′
(11) dx[f(x)g(x)h(x)] = f (x)g(x)h(x)+f(x)g (x)h(x)+f(x)g(x)h(x)
3
(12) d [f(x)] = f′(x)g(x)−f(x)g′(x) Quotient Rule
2
dx g(x) (g(x))
(13) d [f ◦ g(x)] = d [f(g(x)] = f′(g(x))g′(x) Chain Rule
dx dx
Following formulas are special forms of formula (13), but they are most commonly
used forms when you are taking the derivatives of composite functions.
d f(x) f(x) ′
(a) dx[e ] = e f (x)
(b) d [lnf(x)] = 1 f′(x)
dx f(x)
d n n−1 ′
(c) dx[(f(x)) ] = nf(x) f (x)
d p 1 −1 ′
(d) dx[ f(x)] = 2f(x) 2f (x)
d ′ ′ ′
(e) dx[f(g(h(x)))] = f (g(h(x)))g (h(x))h (x)
4
Answers
(1) f′(x) = 2x +2
′ x x
(2) g (x) = 3e +3xe
′ 1
(3) h (x) = 2 (2x+1)
x +x
′ 2 7 7
(4) t (x) = 9x e Here e is just a constant coefficient, has nothing to do with x, so just
keep it.
′ 1·(x−1)−(x+1)·1
(5) n (x) = 2
(x−1)
′ t2 t2
(6) a (t) = 1 · e +te (2t)
u 2 1 u
2uln(1+e )−u ue
(7) f′(u) = 1+e
u 2
(ln(1+e ))
√
4 4 2 1 −3
(8) g′(x) = e 3x +3x +1 · 4(3x4 + 3x2 + 1) 4(12x3 + 6x)
′ −3
(9) h (y) = −2(7y) · 7
2 3 2 1 3x 3 2 3x
′ [(15x +4x)ln(x)+(5x +2x +2)x](e +x)−(5x +2x +2)ln(x)(3e +1)
(10) s (x) = 3x 2
(e +x)
(11) f′(x) = 100(x2 +x)99(2x+1)
(12) g′(x) = (6x + 1)exlnx+(3x2 +x+1)exlnx+(3x2 +x+1)ex1
x
x 2 x 2 x 1
′ [(2x+1)4 +(x +x+1)ln4·4 ]xlnx−(x +x+1)4 [1·lnx+xx]
(13) h (x) = 2
(xlnx)
1 2 2
′ x −x 2 x −x
(14) t (x) = 2 (2x+3)e +ln(x +3x)e (2x−1)
x +3x
′ −2 1
(15) n (x) = −(lnx+x) (x +1)
′ 3 t 2 t
(16) a (t) = 1 · (t + t + e ) + t(3t + 1 + e )
′ 7 −2
(17) f (u) = −(e +ln2+1)u
′ 1 3 −1 2
(18) g (x) = 2(x +2x+1) 2(3x +2)
′ 5y 1 −6
(19) h (y) = 5e +2y2−5y
′ (3t2+7)t−(t3+7t+1)·1
(20) s (t) = 2
t
(21) f′(x) = 1 [ex(2x) + ex(x2 + 1)] or f′(x) = 1 + 1 2x
x 2 2
e (x +1) x +1
2 7 5
′ (x +3) (x−1) 2 6 5 2 7 4
(22) g (x) = e [7(x +3) (2x)(x−1) +(x +3) 5(x−1) ]
′ 1 1 −1
(23) h (x) = 2x can first rewrite ln(x) = ln(x ) = −ln(x)
′ x
(24) t (x) = y(ln2·2 ) here y has nothing to do with x, so think it as a constant coefficient,
just keep it.
′ 3 4 3 4 3 3 4 4 2
(25) n (x) = 4(x +1) (x−1) x +(x+1) 4(x−1) x +(x+1) (x−1) 3x
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