Di erentiation Practice Key with some Hints

Differentiation Practice Key with some Hints

Summer Atkins October 2018

Just want to say that this differentiation Practice is very time consuming. I don't expect students to do all of these problems. The answers that I am providing may be wrong, because to quote the philosopher Cyrus "Everybody makes mistakes". You can always check your problems via wolfram alpha. However, it is possible that your solution could look different from mine because I might have simplified it differently.

1 Problem 1 Answers

(a) f (x) = e + 3x2 + (sin e)ex. Hint: Note that e and sin e are considered constants so use constant rule [cf (x)] = cf (x).

(b) f (x) = 7e2x6 + ln ()x + x-1 Hint: Use [ax] = ln (a)ax for deriving x and power rule for deriving x.

(c) f (x) = (2/3)x-1/3 - (8/15)x-9/5 + (7/6)x-12/6 + 18x or you could write f (x) = 2 - 33x

8

7

+ + 18x.

15 5 x9 6 6 x12

Hint:

rewrite

3 x2 as x2/3 and

3 52x4

as

2 3

x-4/5

and

so

on.

(d)

f

(x) = e cos x +

2 15

e2x-4/3

+

1 25

1 1 + x2 .

See

above

hint

and

note

that

d dx

(tan-1

x)

=

1 1+x2

(e)

f

(x) = 7x6 - csc x cot x -

e x ln (5)

+

2x 9

.

Hint:

You

change

of

base

formula

for

logs

to

change

log5 x

to

ln (x) ln 5

,

then

derive.

2 Problem 2 Answers

You can approach these problems one of two ways, you can rearrange these rational expressions in to looking like products of two terms. With this approach you probably should expand the numerator before trying to rewrite the expression. Another way to derive the functions is that you can use logarithmic differentiation. You will learn logarithmic differentiation after you learn implicit differentiation. You can use these problems to practice quotient rule, but the other methods might cause my solution to be simplified differently. I would definitely study logarithmic differentiation because I am certain that it will come up on the exam.

7x4 + 6x2 - 1

4x(x2 + 1) (x2 + 1)2

(a) f (x) =

or you could say f (x) = -

2 x3

x

2x x

Hint:

If using quotient rule you will need to use chain rule to find

d dx

[(x2

+

1)2].

When

using logarithmic differentiation you can use log rules to where differentiating becomes a

lot easier.

1

(b)

fH(ei(xnxt+): =noext2)e2+th=aetex2(xe3x/2++2xe)x2

= e2x2

+x

+

2ex x

+

x

so

x

(c) f (x) = ex - e-x

3x3 + x2 + 3x + 9

(d) f (x) =

2x2 x

I think the easiest way to go about this one is logarithmic differentiation.

(e) f (x) = (-2/3)x-4/3 - (2/9)x-5/3

3 Problem 3 Answers

I would skip (g) and (i) (or just stop after (f) for this section). problem (h) is fun

(a)

f

(x) =

2 9

sin-1(x)

+

x2 9

1

.

1 - x2

Note:

d dx

sin-1(x)

=

1 1-

x2

(b) f (x) = 3x2ex(3 + x).

(c) f (x) = 9

cos x

- sin x sin-1 x .

1 - x2

Note: sin x sin-1 x = 1!

(d) f (x) = ex sec x(1 + x + x tan x) Hint: [f gh] = f gh + f g h + f gh

(e) f (x) = xex 2 sin-1 x + x sin-1 x + x

see hint (d)

1 - x2

(f )

f

(x)

=

15x4

+

33 2

x9/2

(g)

f

(x) =

ex tan x +xex(tan x+sec x)- 2x

ln (9)9x(sin-1 x log5 x + 9x log5 x

See hint in (d) also use change of base formula for logs.

1

1 - x2

9x sin1 x +

x ln (5)

(h) f (x) = 0 Hint: use some trig identities.

(i) Typo in problem don't worry about it.

(j)

f

(x)

=

(-9)3x[

4 5

x-1/5

sin

x

+

x4/5(cos

x

+

ln

(3)

sin

x]

4 Problem 4 Answers

don't worry about doing all of these problems ( maybe do (a)-(f) if you want to get the hang of things).

ex(x2 - 2x + 1) (a) f (x) = (x2 + 1)2 .

(b)

f

(x) = -

3e2

(x

+

3)2

2x

Note: e2 is a fixed number so (e2) = 0.

2

6x - 3 sin-1 x 1 - x2

(c)

f

(x) =

2x

-

x3(sin-1

(x)

+

x)2

Note:

(sin-1 x)

=

1 1-x2

-6 csc x - 2 - 6 csc x cot x sin-1(x) 1 - x2

(d) f (x) =

1 - x2(3 csc x + 1)

3ex

+

3 x(sec

x)(tan

x)ex

-

6xex

-

3 x(sec

x)ex

(e) f (x) =

9 xe2x

(f )

f

(x) =

x(cos

x

+

ex)

-

2

+

1 2x

(ex

-

sin

x)

( x - 2 cos x)2

Hint: cos2 x + sin2 x = 1

(2x - 3)(9x + sec x) - (x2 - 3x + 2)(ln (9)9x + sec x tan x

(g) f (x) =

(9x + sec x)2

Hint: see hint 1(b).

(h)

f

(x) =

-3 1-x2

-

1 2x

9 tan-1 x

3

cos-1

x

-

x

- 9(1 - x2)(tan-1 x)2

Hint

d dx

cos-1(x)

=

-1

1 - x2

and

d dx

tan-1

x

=

1 1+x2

(i)

f

(x) =

(3

-

1 2x

)(e2

-

2

sin

x)

-

(3x

-

x)(ex

-

2

cos

x)

(ex - 2 sin x)2

(j) typo so don't worry about it.

5 Problem 5 Answers

Don't do (g)-(j) they look terrible. Those might be easier with logarithmic differentiation though

(1 - x)ex sin x - xex cos x

(a) f (x) =

(ex sin x)2

1

x tan x

1

(b) f (x) = 2

x(cos

x

ln

x

+

cos

x

ln x

+

x cos x(ln x)2

(c)

f

(x) =

log3

x

+

1 ln 3

csc x - 1

+

x

log3 x(csc (csc x -

x cot 1)2

x

(d)

f

(x) =

ex

cos

x(ln

x(1

+

x

-

1 2x

)

-

1 x

(1

+

x))

-

ex

sin

x

(ln (x)(1 + x))2

3x

12

ln

(3) x

cot

x

tan-1

x

+

12 1

x cot + x2

x

-

tan-1

x(

6 x

cot

x

-

12 x

csc2

x)

(e) f (x) =

( x + 3)2

ex

x ln x

(ln x + x ln x + 1)( x + 3) -

(f) f (x) =

2x

( x + 3)2

Note deriving the numerator requires applying product rule twice.

(xex ln x) = x ex ln x + x(ex) ln x + xex(ln x) = ex ln x + xex ln x + xex(1/x) = ex(ln x + x ln x + 1)

3

6 Problem 6 Answers

(a) f (x) = esin x cos x

9tan-1 x ln (9) (b) f (x) = 1 + x2 (c) f (x) = (2 ln 3)(sin x)(cos x)3sin2 x

Gonna be using chain rule twice:

(3sin2 x) = 3sin2 x ln (3) ? (sin2 x) = 3sin2 x ln (3) ? 2(sin x) ? (sin x) = 3sin2 x ln (3) ? 2(sin x) ? (cos x).

1 (d) f (x) = - 1 + x2

-2 cos x sin x - sec2 x (e) f (x) = (cos2 x - tan x)

Gonna be using chain rule more than once:

(ln (cos2 x - tan x))

=

(cos2

x

1 -

tan

x)

?

(cos2

x

-

tan

x)

=

1 (cos2 x - tan x) ? (2 cos x ? (cos x)

- sec2 x)

=

(cos2

x

1 -

tan x)

?

(2 cos x

?

(- sin x)

-

sec2

x)

(f) f (x) = -x(csc x2)(cot x2) (ecsc x2 )

(ecsc x2

= 1

? (ecsc x2 )

2 ecsc x2

1 =

? (ecsc x2 ) ? (csc x2)

2 ecsc x2

1 =

? (ecsc x2 ) ? (- csc x2 cot x2) ? (x2)

2 ecsc x2

1 =

? (ecsc x2 ) ? (- csc x2 cot x2) ? (2x)

2 ecsc x2

4

Note you can also use logarithmic differentiation so that you only use chain rule once!

y = ecsc x2

ln y = ln ( ecsc x2 )

take ln of both sides

ln y = 1 ln (ecsc x2 ) using log rule 2

ln y = 1 csc (x2) using log rule 2

(ln y) = 1 csc (x2) 2

1 y

=

1 (- csc x2 cot x2) ? (x2)

implicit diff. on LHS

y2

1 y

=

1 (- csc x2 cot x2) ? (2x)

y2

y = y ? (-x csc (x2) cot (x2))

y = ( ecsc x2 )(-x csc (x2) cot (x2))

-5 tan x sec2 x (g) f (x) =

(5 tan2 x)3 (h) f (x) = (tan3 x)(3 sec x + cos x)

7 Problem 7 Answers

(a) f (x) = (ln 10)10(x cot x)(cot x - x csc2 x) Using chain rule then product rule when deriving inside function

(b) f (x) = x8ex sin x(9 + x sin x + x2 cos x) Use product rule first, then use chain rule when finding (ex sin x) , then use product rule again when finding (x sin x)

(c) f (x) = sec ex(cot2 x) tan ex(cot2 x) ex(cot2 x) [cot2 x - 2(csc2 x)(cot x)] First use chain rule (note: ex(cot2 x) is our inside function). Second, use chain rule again to compute (ex(cot2 x)) where the inside function is x(cot2 x). Third use product rule to find (x(cot2 x)) and you will need chain rule to compute (cot2 x) = 2(cot x) ? (cot x) .

(d) meh

(e) meh

8 Problem 8 Answers

I am not concerned if you skip these problems. (a) f (x) = (b) f (x) = (c) f (x) = (d) f (x) = (e) f (x) =

5

9 Problem 9 Answers

Not too concerned if you skip these problems (a) f (x) = (b) f (x) = (c) f (x) = (d) f (x) = (e) f (x) = (f) f (x) =

10 Problem 10 Answers

2x ln (2) - 2ex - y2 (a) y =

2xy

Steps for solving problem:

y2x + 2ex = 2x

2y dy x + y2 + 2ex = 2x ln (2) use product rule on LHS! dx

dy 2xy

=

2x ln (2) - y2

- 2ex

dy solve for

dx

dx

dy 2x ln (2) - y2 - 2ex

=

dx

2xy

(b)

dy dx

=

2x2

4xy - 9y2e9y + 2y2ey + 81xy2e9y

Steps for solving the problem:

2x2 - 2ey = 9xe9y y

4xy

-

2x2

dy dx

y2

- 2ey dy dx

=

9e9y

+ 9xe9y(9 dy ) dx

use quotient rule for LHS and product rule for RHS

4xy - 2x2 dy - 2y2ey dy = 9y2e9y + 9xy2e9y(9 dy ) multiply everything by y2

dx

dx

dx

4xy - 2x2 dy - 2y2ey dy = 9y2e9y + 81xy2e9y dy

dx

dx

dx

4xy - 9y2e9y = 2x2 dy + 2yey dy + 81xy2e9y dy moving terms with dy/dx to the LHS

dx

dx

dx

4xy - 9y2e9y = (2x2 + 2yey + 81xy2e9y) dy dx

dy

4xy - 9y2e9y

dx = 2x2 + 2y2ey + 81xy2e9y .

(c)

Skip

(method

should

be routine

but we have to convert

3 log5 y =

3 ln y ln 5

).

6

(d)

dy dx

=

(ln

9)9x

ln

y (ln

y)

+

x y

1 y

+

ln x

-

(ln 9)9x ln y

x y

Note for these problems you can just write y

instead of

dy dx

.

0 = ln y + y ln x - 9x ln y

1 0= y +

1 y ln x + y

- (ln 9)9x ln y ? (x ln y)

(chain and product rule)

y

x

y

y

0 = + (ln x)y +

- (ln 9)9x ln y ?

x 1 ln y + y

y

x

y

(product rule)

0 = 1 + ln x - (ln 9)9x ln y x y + x - (ln 9)9x ln y(ln y) grouping y' terms together

y

y

y

(ln 9)9x ln y(ln y) + x = 1 + ln x - (ln 9)9x ln y x y

yy

y

(ln

9)9x ln y(ln

y)

+

x y

1 y

+

ln x

-

(ln

9)9x

ln

y

x y

=y

moving non y' terms to LHS

y

(e)

dy dx

=

3 y - 1 + x2

1

-

3x

+

1

(tan-1 x)

3 3 y2 2 y 2 y

Steps for solving problem: First let's change roots into fractional exponents and then

derive.

y1/3 - 3y1/2x + y1/2 tan-1 x = 3

1 y-2/3y - 3 3

1 y-1/2y x + y1/2 2

+

1 y-1/2y 2

tan-1

x

+

y1/2

1

1 +

x2

=0

1 y-2/3y 3

- 3 xy-1/2y 2

- 3y1/2 + 1 y-1/2(tan-1 x)y 2

y1/2 + 1 + x2

=0

1

-

3x

+

1

(tan-1 x)

3 3 y2 2 y 2 y

y

y = 3 y - 1 + x2

y

y=

1

3 y - 1 + x2

-

3x

+

1

(tan-1 x)

3 3 y2 2 y 2 y

(f )

dy dx

=

Steps for solving first change roots into fractional exponents and use exponent rules, then

7

derive.

y3/2 - 2y1/2y1/3 - y1/2x1/2y = 0

y3/2 - 2y5/6 - x1/2y3/2 = 0

3 y1/2y - 5 y-1/6y - 1 x-1/2y3/2 - 3 x1/2y1/2y = 0

2

3

2

2

3 y1/2 - 5 y-1/6 - 3 x1/2y1/2

2

3

2

3

5 3

y- - x y

2

36y 2

y = 1 x-1/2y3/2 2 yy

y= x

y y

y

=

3 2

y

-

x

3 56 y -

3 2

xy

yy

y

=

x(

3 2

y

-

3 56 y

-

3 2

xy)

yy

y

=

3 2

xy

-

5 3

6 xy

-

3 2

xy

11 Problem 11 Answers

I would have said that these problems would be good to look at but most of these problems are written badly to where I can't tell what the exponent is. Porblems (b),(c), (k)-(n) are more of what would be expected of you to know. If you are crunch for time I would study problems (k)(n).

(a) Skip (a) because I can't tell if whether the function is (sin x)log3 x of (sin (xlog3 x))

(b) y = (ln x)tan-1 x

ln (ln x) tan-1 x +

1 + x2 x ln x

Steps for solving problem:

y = (ln x)tan-1 x

ln y = ln (ln x)tan-1 x

ln y = (tan-1 x) ln (ln x)

1 y

= (tan-1 x)

ln (ln x) + (tan-1 x)(ln (ln x))

y

y y

=

1

1 + x2

ln

(ln

x)

+

(tan-1

x)

1 ? (ln x)

ln x

y y

=

ln (ln x) 1 + x2

+

(tan-1

x)

11 ?

ln x x

y ln (ln x) tan-1 x

= y

1 + x2

+

x ln x

ln (ln x) tan-1 x y = y 1 + x2 + x ln x

y = (ln x)tan-1 x

ln (ln x) tan-1 x +

.

1 + x2 x ln x

(take ln of both side) (using log rules for exponents)

(using product rule)

(using chain rule on last term)

(c) y = (ex)tan x[tan x + sec2 x]

8

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