Derivatives Test Problems - Calculus I | MATH 131, Exams of Calculus

Material Type: Exam; Class: Calculus I; Subject: Mathematics; University: University of Massachusetts - Amherst; Term: Spring 2003;

Typology: Exams

Pre 2010

Uploaded on 08/18/2009

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Derivatives Test (Spring 2003) MATH 131
We compute the derivative and do not simplify:
d
dx 5x64x2+ 3x3/22= 30x5+ 8x3+ 33
2x1/2
(1)
d
dx (sec(x)) = sec(x)·tan(x)(2)
d
dx
cos(x)
x3+ex=(sin x)(x3+ex)(cos x)(3x2+ex)
(x3+ex)2
(3)
d
dx ((5x2+ 2x+ 3) ·ex) = (10x+ 2) ·ex+ (5x2+ 2x+ 3) ·ex
(4)
d
dx (esin(x)) = esin(x)cos(x)(5)
d
dx (tan(ex)) = sec2(ex)·ex
(6)
d
dx ((2x23x2)·sin(x3+ 1)) =(7)
= (4x+ 6x3)·sin(x3+ 1) + (2x23x2)·(cos(x3+ 1) ·(3x2))
d
dx (4x223x+ 1)4=4(4x223x+ 1)5·(8x21
2(3x+ 1)1/2·3)(8)
d
dx cos3(x) + ex2
x3= 3 cos(x)·(sin(x)) + ex2
·(2x)3x2
(9)
d
dx sin x2ex = cos x2ex·
1
2(x2ex)1/2·(2xex)(10)

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Derivatives Test (Spring 2003) MATH 131

We compute the derivative and do not simplify:

d dx

5 x^6 − 4 x−^2 + 3x^3 /^2 − 2

= 30x^5 + 8x−^3 + 3

(1) x^1 /^2

d dx

(2) (sec(x)) = sec(x) · tan(x)

d dx

cos(x) x^3 + ex^

(− sin x)(x^3 + ex) − (cos x)(3x^2 + ex) (x^3 + ex)^2

d dx (4) ((5x^2 + 2x + 3) · ex) = (10x + 2) · ex^ + (5x^2 + 2x + 3) · ex

d dx

(5) (esin(x)) = esin(x)^ cos(x)

d dx (6) (tan(ex)) = sec^2 (ex) · ex

d dx

(7) ((2x^2 − 3 x−^2 ) · sin(x^3 + 1)) =

= (4x + 6x−^3 ) · sin(x^3 + 1) + (2x^2 − 3 x−^2 ) · (cos(x^3 + 1) · (3x^2 )) d dx

(4x^2 − 2

3 x + 1)−^4

= −4(4x^2 − 2

3 x + 1)−^5 · (8x − 2

(8) (3x + 1)−^1 /^2 · 3)

d dx

cos^3 (x) + ex 2 − x^3

= 3 cos(x) · (− sin(x)) + ex 2 (9) · (2x) − 3 x^2

d dx

sin

x^2 − ex

= cos

x^2 − ex

(10) (x^2 − ex)−^1 /^2 · (2x − ex)