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7/29/2019 Deriv Probs
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Derivatives - Problems and Solutions
Laurel Benn
January 15, 2010
This problem and solution set should be an invaluable resource to dealwith the problems we have encountered in derivatives so far. I urge you tospend more time reading your notes, going over the problems we worked inclass, and working additional problems. Remember there is no substitutefor hard work.
Example 1 If f (x) =√
5x2 + 1, use the definition of the derivative to find
f ′(x).
Solution: f ′(x) = limh→0
f (x + h) − f (x)
h= lim
h→0
5(x + h)2 + 1 −
√ 5x2 + 1
hNext multiply both numerator and denominator by
5(x + h)2 + 1 +√
5x2 + 1 to get
limh→0
5(x + h)2 + 1 − (5x2 + 1)
h(
5(x + h)2 + 1 +√
5x2 + 1)= lim
h→0
5(x2 + 2xh + h2) + 1 − 5x2 − 1
h(
5(x + h)2 + 1 +√
5x2 + 1)
= limh→0
10xh + h2
h(
5(x + h)2 + 1 +√
5x2 + 1) = limh→0
10x + h 5(x + h)2 + 1 +
√ 5x2 + 1
= 10x2√ 5x2+1
= 5x√ 5x2+1
For other similar problems the setup is the same but we may have to do
a different algebraic manipulation to get what we want.
In the next example, I am assuming that we remember from class that(ln x)′ = 1
x.
Example 2 Find the derivative of f (x) = log7 x.
Solution:
f (x) = log7 x = ln xln 7 , by the change of base formula.
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⇒ f ′(x) = 1ln 7
(lnx)′, since ln 7 is a constant.
⇒ f ′(x) = 1ln 7
· 1x
= 1x ln 7
In fact for any base a, (loga x)′ = 1x ln a
.
In class I showed you another way to solve this type of problem. Please refer
to your class notes.
Example 3 Find the derivative of y = x3 ln 2xex sinx
Solution:Take natural log of both sides of the equation to get:
ln y = ln(x3 ln 2x
ex sinx)
⇒ ln y = ln(x3 ln 2x) − ln(ex sin x)
⇒ ln y = lnx3 + ln(ln 2x) − ln ex − lnsinx
⇒ 1ydydx
= 3x2
x3+ 1
x ln 2x− 1 − cosx
sin x
⇒ 1
y
dy
dx=
3
x+
1
x ln 2x− 1 − cotx
⇒ dy
dx= y(
3
x+
1
x ln 2x− 1 − cotx)
⇒ dy
dx=
x3 ln 2x
ex sin x(
3
x+
1
x ln 2x− 1 − cotx)
This problem could also have been done with the quotient rule.
Example 4 Find the derivative of y = e2x cos3x√ (x−4)
Solution:
Take natural log of both sides of the equation to obtain:
ln y = ln e2x + ln cos3x− ln(√ x− 4)
⇒ ln y = 2x + ln cos3x− 12 ln(x− 4).
Differentiating both sides of the equation we get:1y
dy
dx= 2 + (−3sin3x)
cos3x− 1
21
x−4⇒ dy
dx= y(2 − 3tan3x− 1
2(x−4)
⇒ dy
dx= e2x cos3x√
x−4(2 − 3tan3x− 1
2(x−4)).
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Example 5 Find the derivative of y =√
3x2 + 4x− 1.
y =√
3x2 + 4x− 1 = (3x2 + 4x− 1)1
2
dy
dx=
1
2(3x2 + 4x− 1)−
1
2 (6x + 4) =3x + 2√
3x2 + 4x−
1
Example 6 Find the derivative of y = 5x4 + 4x− 12x2 + 1√
x− 3.
y = 5x4 + 4x− x−2
2+ x−
1
2 − 3
dy
dx= 20x3 + 4 + x−3 − 1
2x−
3
2
= 20x3 +1
x3− 1
2√ x3
+ 4
Example 7 Find the derivative of f (x) = ln x3 .
f ′(x) = (ln x3)′ = (3ln x)′ = 3x
.
Example 8 If f (x) = ln[ln(x4 + 1)]. Find f ′(x).
Using the chain rule we get,
f ′(x) =1
ln(x4 + 1)· (ln(x4 + 1))′
=1
ln(x4 + 1)· 1
x4 + 1· 4x3
= 4x3
(x4 + 1) ln(x4 + 1)
Example 9 For the function f (x) = e−x2
2 , compute the first, second and
third derivatives of f (x).
f ′(x) = e−x2
2 · −2x
2= −xe−
x2
2
f ′′(x) = −x(−xe−x2
2 ) + e−x2
2 · (−1)
⇒ f ′′(x) = e−x2
2 (x2 − 1) = (x− 1)(x + 1)e−x2
2
f ′′′(x) = 2xe−x2
2 + −xe−x2
2 (x2 − 1) = x(3 − x2)e−x2
2
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Example 10 Compute the derivative of y = 4π2.
Since π is a constant ⇒ 4π2 is a constant.
⇒ dy
dx= 0
Example 11 Differentiate y = sin x + 10 tanx w.r.t x.d
dx(sinx + 10 tanx) = cos x + 10sec2 x
Example 12 Differentiate y = xcosx
.
By the quotient rule we get,
x
cosx
′=
cosx(1) − x(− sin x)
cos2 x=
cosx + x sin x
cos2 x
Example 13 Differentiate y = sin(x cos x).
dy
dx = cos(x cos x) · (cosx− x sinx)
Example 14 For f (x) = ex−e−xex+e−x
. Find f ′(x).
Using the quotient rule we get ,
f ′(x) =(ex + e−x)(ex + e−x) − (ex − e−x)(ex − e−x)
(ex + e−x)2
=(ex + e−x)2 − (ex − e−x)2
(ex + e−x)2
=(ex + e−x + ex − e−x)(ex + e−x − ex + e−x)
(ex + e−x)2
=(2ex)(2e−x)
(ex + e−x)2
=4
(ex + e−x)2
Example 15 Find dydx
for y =√
1 + x1234.
dy
dx=
1
2√
1 + x1234· (1234x1233)
=
617x1233
√ 1 + x1234
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Example 16 For f (x) = x ln x− x. Find f ′(x).
f ′(x) = ln x + x
1
x
− 1
= ln x
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