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Assignment #10— Derivatives of Logarithmic and Exponential Functions
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2003 Doug MacLean
Find f (x) and f (x) if:
x2 + 1
(1) f (x) = ln
x3 − 1
Solution:
f (x) = ln x 2 + 1 − ln x 3 − 1 ,
2
3
3x 2
2x x 3 − 1 − 3x 2 x 2 + 1
2x 4 − 2x − 3x 4 − 3x 2
1
1
2x
x +1 − 3
x −1 = 2
− 3
=
=
=
so f (x) = 2
x +1
x −1
x +1 x −1
(x 2 + 1) (x 3 − 1)
(x 2 + 1) (x 3 − 1)
x 3 + 3x + 2
−x 2
(x + 1) (x 3 − 1)
3x 2
2x
−
:
x2 + 1 x3 − 1
(2x) x 2 + 1 − 2x x 2 + 1
To find f (x), we differentiate f (x) =
2x
3x 2
f (x) =
−
=
2
3
3 x + 1 2 x2 − 1
(6x) x − 1 − 3x (3x )
=
2
(x 3 − 1)
2x 2 + 2 − 4x 2 6x 4 − 6x − 9x 4
−
=
2
2
(x 2 + 1)
(x 3 − 1)
6x + 3x 4
2 − 2x 2
+
2
2 =
(x 2 + 1)
(x 3 − 1)
2
1 − x2
(x 2 + 1)
2
+ 3x
2 + x3
(x 3 − 1)
2
(x 2 + 1)
2
−
3x 2
x 3 − 1 − 3x 2 x 3 − 1
(x 3 − 1)
2
=
2 x 2 + 1 − 2x(2x)
(x 2 + 1)
2
−
2
Solution:
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sin x
(2) f (x) = ln
sin x + cos x
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f (x) = ln (sin x) − ln (sin x + cos x), so
cos x(sin x + cos x) − (cos x − sin x) sin x
1
1
cos x cos x − sin x
−
=
=
(sin x) −
(sin x + cos x) =
sin x
sin x + cos x
sin x sin x + cos x
sin x(sin x + cos x)
sin x cos x + cos2 x − sin x cos x + sin2 x
=
sin x(sin x + cos x)
f (x) =
1
sin x(sin x + cos x)
−2
f (x) = (−1) (sin x(sin x + cos x)) (sin x(sin x + cos x)) =
1
2 ((sin x) (sin x + cos x) + sin x(sin x + cos x) ) =
(sin x(sin x + cos x))
1
((cos x)(sin x + cos x) + sin x(cos x − sin x)) =
2
sin x(sin x + cos x)2
1
2
2
sin
x
cos
x
+
cos
x
+
sin
x
cos
x
−
sin
x
=
sin2 x(sin x + cos x)2
1
2 sin x cos x + cos2 x − sin2 x =
2
2
sin x(sin x + cos x)
π
sin
2x
+
sin 2x + cos 2x
4
=
2
2
2
2
sin x(sin x + cos x)
sin x sin x + π4
3
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(3) f (x) = ln(sec x + tan x)
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Solution:
f (x) =
1
1
(sec x + tan x) =
(sec x tan x + sec2 x) =
sec x + tan x
sec x + tan x
f (x) = sec x tan x
(4) f (x) = x a − ax , where a is a constant.
Solution:
f (x) = ax a−1 − (ln a)ax ,
f (x) = a(a − 1)x a−2 − (ln a)2 ax
sec x
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(5) f (x) = loga x + logx a, where a is a constant.
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2003 Doug MacLean
Solution:
f (x) =
1
x
ln a
f (x) =
ln a
ln x
ln x
+
=
+ ln a(ln x)−1 , so
ln a ln x
ln a
+ ln a(−1)(ln x)−2 (ln x) =
1
ln a
1
1
− ln a(ln x)−2
=
−
=
x ln a
x
x ln a x(ln x)2
1
f (x) =
− ln a x −1 (ln x)−2 =
x ln a
−1
− ln a x −1 (ln x)−2 − ln a x −1 (ln x)−2 =
2
x ln a
−1 −1
−2
−2
−3
−
ln
a
−x
−
ln
a
x
(ln
x)
(−2)(ln
x)
(ln x) =
x 2 ln a
ln a
2 ln a
−1
+
+
x 2 ln a x 2 (ln x)2 x 2 (ln x)3
(ln x)2 − (ln a)2
x ln a(ln x)2
5
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2 −4x+5
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(6) f (x) = 2x
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2003 Doug MacLean
Solution:
f (x) = e(x
2 ln 2(x − 2)2x
and
, so f (x) = e(x
2 −4x+5) ln 2
(x 2 − 4x + 5) ln 2
x 2 −4x+5
x 2 −4x+5
f (x) = 2 ln 2 (x − 2)2
= 2 ln 2 (x − 2) 2
2
2
2 ln 2 (1)2x −4x+5 + (x − 2) 2 ln 2(x − 2)2x −4x+5 =
2
x −4x+5
=
+ (x − 2) 2
2
2 ln 2 1 + 2 ln 2(x − 2)2 2x −4x+5
(7) f (x) = x ln x
Solution:
f (x) =
1
x
= e(x
2 −4x+5
2 −4x+5) ln 2
1
=
f (x) = (x) ln x + x(ln x) = (1) ln x + x
x
ln x + 1
2 −4x+5) ln 2
((2x − 4) ln 2) =
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(8) f (x) = x 2x
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2003 Doug MacLean
Solution:
f (x) = e2x ln x ,
1
so f (x) = e
(2) ln x + (2x)
= x 2x [2 ln x + 2] = 2(ln x + 1)x 2x
(2x ln x) = e
[(2x) ln x) + (2x)(ln x) ] = x
x
2x 1
1
2x
2x
2x
2
+ 0 x +2(ln x+1) 2(ln x + 1)x
+ 2(ln x + 1) x 2x
and f (x) = 2(ln x+1) x +2(ln x+1) x
=2
= 2
x
x
2x ln x
2x ln x
2x
(9) f (x) = x sin x
Solution:
f (x) = e
sin x ln x
, so f (x) = e
sin x ln x
(sin x ln x) = e
sin x ln x
[(sin x) ln x + sin x(ln x) ] = x
sin x
x
cos x ln x +
x
1
1 and f (x) = x sin x
cos x ln x + (sin x)
+ x sin x cos x ln x + (sin x)
=
x
x
1 2
1
1
−1
x sin x cos x ln x + (sin x)
=
+ x sin x − sin x ln x + cos x + (cos x) + sin x
x
x
x
x2
sin x 2
sin x
cos x
sin x
cos x ln x +
−
x
− sin x ln x + 2
x
x
x2
sin x
sin x
1
cos x ln x + (sin x)
x
=
7
Solution:
f (x) = x
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− ln x
=e
−(ln x)2
, so f (x) = e
−(ln x)2
2 −(ln x)
Thus f (x) > 0 on (0,1) and f (x) < 0 on (1, ∞).
ln x −(ln x)2
ln x −(ln x)2 f (x) = −2
e
e
−2
=
x
x
(ln x) x − ln x(x)
ln x −(ln x)2
ln x
−(ln x)2
−2
e
=
−2
e
−
2
x2 x
x
1
x − ln x(1)
(ln x)2 −(ln x)2 −(ln x)2
e
=
+
4
e
−2 x
x2
x2
(ln x)2
2 2(ln x)2 + ln x − 1 −(ln x)2
1 − ln x
−(ln x)2
+4
=
e
e
−2
x2
x2
x2
1
1 √
−1 ± 3
= −1, , or x = , e,
2
e
4 √
1
1 √
∪
, e .
so we have f (x) > 0 on 0,
e, ∞ , and f (x) < 0 on
e
e
Thus f (x) = 0 if ln x =
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ln x
1
(10) f (x) =
x
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=e
−(ln x)2
1
−2 ln x
x
=
−2
ln x −(ln x)2
e
.
x