taller 12 lhopital der de fcs inversas
DESCRIPTION
lhopital derivadas de funcion inversaTRANSCRIPT
7/21/2019 TALLER 12 LHopital Der de Fcs Inversas
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lımx→π
sin x2 + cos x
1 + sin2 x + cos x
00
lımx→π
sin x
2 + cos x
1 + sin2 x + cos x= LH
lımx→π
12 cos x2 − sin x
2sin x cos x − sin x
= lımx→π
12 cos x2 − sin x
sin2x − sin x
lımx→π
12 cos x2 − sin x
sin2x − sin x = LH
lımx→π
−14 sin x
2 − cos x
2cos2x − cos x
= −1
4 + 1
2 + 1
= 1
4
f (t)
t
f (t) = A
c2 − k2 (sin(kt) − sin(ct))
A,c,k c = k c k
7/21/2019 TALLER 12 LHopital Der de Fcs Inversas
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lımc→k
A
c2 − k2 (sin(kt) − sin(ct)) =
LH
lımc→k
−A
2c · t cos(ct)
= −At cos(kt)
2k
lımx→0
x1
ln(exp(x)−1)
v = x
1ln(exp(x)−1)
ln v = 1
ln(exp(x) − 1) · ln x
lımx→0
ln v = lımx→0
ln x
ln(exp(x) − 1)
ln
lımx→0
v
= lımx→0
ln x
ln(exp(x) − 1)
ln
lımx→0
v
= lımx→0
1
x
exp(x)exp(x)−1
= lımx→0
exp(x) − 1
x · 1
exp(x)
= 1
ln
lımx→0
v
= 1
lımx→0
v = e
lımx→+∞
x1x
L = lımx
→+
∞
x1x
ln L = lımx→+∞
ln(x)x
7/21/2019 TALLER 12 LHopital Der de Fcs Inversas
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ln L = lımx→+∞
1
x= 0
L = e0 = 1
, f (x) = lnx + √ x2 + 1 + arctan1−x21+x2
f (x)
f (x) = 1
x +√
x2 + 1·
1 + 1
2√
x2 + 1· 2x
+
1
1 +
1 − x2
1 + x2
2 ·−2x(1 + x2) − (1 − x2) · 2x
(1 + x2)2
= x +
√ x2 + 1
x +√
x2 + 1· 1√
x2 + 1+
(1 + x2)2
(1 + x2)2 + (1 − x2)2 ·−2x · 1 + x2 + 1 − x2
(1 + x2)2
=
1
√ x2 + 1 − 4x
1 + 2x2 + x4 + 1 − 2x2 + x4
= 1√
x2 + 1− 4x
2 + 2x4
= 1√
x2 + 1− 2x
1 + x4
f (x) = arctan(x) + arctan1x
f (x) x ∈ Dom(f )
f (x) = 1
1 + x2 + 1
1 + 1x2
·− 1
x2=
1
1 + x2 − x2
x2(1 + x2)
= 1
1 + x2 − 1
1 + x2
= 0
f (x) = arctan√
x2 − 1 f (x) = d
dx ( (x)) x > 0
7/21/2019 TALLER 12 LHopital Der de Fcs Inversas
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f (x) = 1
1 + x2 − 1 · 1
2√
x2 − 1· 2x
= x
x2√
x2 − 1
= 1
x√
x2
−1
= d
dx ( (x))
f (x) = arcsin√
1 − x2 |x| < 1
f (x) =
d
dx (arcsin(x)) , −1 < x < 0
d
dx (arc cos(x)) , 0 < x < 1
f (x) = 1 1 − (1 − x2)
· 12√
1 − x2 · (−2x)
= − 1√ x2
x√ 1 − x2
= − x
|x|√ 1 − x2
−1 < x < 0 |x| = −x
f (x) = 1√
1 − x2
= ddx
(arcsin(x))
0 < x < 1 |x| = x
f (x) = − x
x√
1 − x2
= − 1√ 1 − x2
= d
dx (arc cos(x))
f (x) =
d
dx (arcsin(x)) , −1 < x < 0
d
dx (arc cos(x)) , 0 < x < 1
f (x) = 1 − 4 arcsin2(2x)
X
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X
f (x) = 0
f (x) = −4 · 2 arcsin(2x) · 1√ 1 − 4x2
· 2
= −16 arcsin(2x)√ 1 − 4x2
f (x) = 0
−16 arcsin(2x)√ 1 − 4x2
= 0
arcsin(2x) = 0
x = 0
F (x) f (x)
F (x) = f (x)
F (x) =
√ 1 − x2 + x arcsin(x)
f (x) = arcsin(x)
F (x) = 1
2√
1 − x2 · (−2x) + arcsin(x) + x · 1√
1 − x2
= − x√ 1 − x2
+ arcsin(x) + x√ 1 + x2
= arcsin(x)
= f (x)
f y = f (x) = sin(b arcsin(x)) b ∈ R f
(1 − x2)y − xy + b2y = 0
y = cos (b arcsin(x)) · b√ 1
−x2
= b cos(b arcsin(x))√
1 − x2
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y = b
− sin(b arcsin(x)) · b√ 1−x2 ·
√ 1 − x2 − cos(b arcsin(x)) · 1
2√ 1−x2 · (−2x)
1 − x2
= b
−b sin(b arcsin(x)) + cos(b arcsin(x)) · x√ 1−x2
1 − x2
(1 − x2)y = (1 − x2) · b
−b sin(b arcsin(x)) + cos(b arcsin(x)) · x√ 1−x2
1 − x2
= −b2 sin(b arcsin(x)) + bx cos(b arcsin(x))√
1 − x2
−xy = −x · b cos(b arcsin(x))√ 1 − x2
= −bx cos(b arcsin(x))√ 1−
x2
b2y = b2 sin(b arcsin(x))
(1 − x2)y − xy + b2y = −b2 sin(b arcsin(x)) + bx cos(b arcsin(x))√
1 − x2 − bx cos(b arcsin(x))√
1 − x2 + b2 sin(b arcsin
= 0
f