derivada implicita- Jessica

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Derivadas Impĺıcita
Calculo I
May 12, 2021
6 Calcular derivada implicita
a)x2 + xy + y2 − 5x− 3y = 2
dy
dx
=?
d
dx
(x2 + xy + y2 − 5x− 3y) = d
dx
(2)
d
dx
(x2) +
d
dx
(xy) +
d
dx
(y2)− d
dx
(5x)− d
dx
(3y) = 0
d
dx (xy) =
d
dx (x)y + x
d
dx (y) = y + x
d
dx (y)
d
dx (y
2) = (y2)′ ddx (y) = 2y
d
dx (y)
2x+ y + x
d
dx
(y) + 2y
d
dx
(y)− 5− 3 d
dx
(y) = 0
+x
d
dx
(y) + 2y
d
dx
(y)− 3 d
dx
(y) = −2x− y + 5
(+x+ 2y − 3) d
dx
(y) = −2x− y + 5
dy
dx
=
−2x− y + 5
(+x+ 2y − 3)
b)ln (xy ) = 1
ln
(
x
y
)
= ln (x)− ln (y)
d
dx
(ln (x)− ln (y)) = d
dx
(1)
d
dx
(ln (x))− d
dx
(ln (y)) = 0
1
1
x
− 1
y
dy
dx
= 0
1
y
dy
dx
=
1
x
dy
dx
=
y
x
c) y cos (2x) + sen2(y) = y
d
dx
(y cos (2x) + sen2(y)) =
dy
dx
d
dx
(y cos (2x)) +
d
dx
(sen2(y)) =
dy
dx
d
dx (y cos (2x)) =
dy
dx (cos (2x)) + y
d
dx (cos (2x)) =
dy
dx (cos (2x))− 2y(sen(2x))
d
dx (sen
2(y)) = sen(y) cos (y) dydx + sen(y) cos (y)
dy
dx = 2sen(y) cos (y)
dy
dx
dy
dx
(cos (2x))− 2y(sen(2x)) + 2sen(y) cos (y)dy
dx
=
dy
dx
dy
dx
(cos (2x)) + 2sen(y) cos (y)
dy
dx
− dy
dx
= 2y(sen(2x))
dy
dx
(cos (2x)) + 2sen(y) cos (y)− 1) = 2y(sen(2x))
dy
dx
=
2y(sen(2x))
(cos (2x)) + 2sen(y) cos (y)− 1)
d)y2 = e−
1
x
d
dx
y2 =
d
dx
(e−
1
x )
2y
dy
dx
= (− 1
x
)′(e−
1
x )
dy
dx
=
1
2y
1
x2
(e−
1
x ) =
1
2yx2
(e−
1
x )
7 Derivadas impĺıcita e função inversa
b)y = arccos (ln (x))
2
cos (y) = ln (x)
d
dx
(cos (y)) =
d
dx
(ln (x))
−sen(y)dy
dx
=
1
x
dy
dx
= − 1
xsen(y)
sen2(y) + cos2(y) = 1
dy
dx
= − 1
x
√
(1− cos2(y))
dy
dx
= − 1
x
√
(1− cos2(arccos (ln (x))))
dy
dx
= − 1
x
√
1− (ln (x))2
a)y = arccossec(sec (x))
cossec(y) = sec (x)
(cossec(y))′ =
(
1
sen(y)
)′
= − cos (y)sen2(y) = −cotg(y)cossec(y)
d
dx
(cossec(y)) =
d
dx
(sec (x))
−cotg(y)cossec(y)dy
dx
= tg(x) sec (x)
dy
dx
= − tg(x) sec (x)
cotg(y)cossec(y)
dy
dx
= − tg(x) sec (x)
cotg(arccossec(sec (x)))cossec(arccossec(sec (x)))
dy
dx
= − tg(x) sec (x)
cotg(arccossec(sec (x)))| sec (x)|
dy
dx
= − tg(x) sec (x)
cotg(arccossec(sec (x)))| sec (x)|
cotg2(t) + 1 = cossec2(t)
dy
dx
= − tg(x) sec (x)√
cossec2(arccossec(sec (x)))− 1| sec (x)|
dy
dx
= − tg(x) sec (x)√
(sec2 (x)− 1)| sec (x)|
3
Exerćıcio 4
w)f(s) = (a+ bs)ln(a+bs) y = eln(y)
f(s) = eln(a+bs)
ln(a+bs)
f(s) = eln(a+bs) ln(a+bs)
f(s) = g(h(s))
g′(t) = et e t = h(s) = (ln(a+ bs) ln(a+ bs))
h′(s) = b(a+bs) ln(a+ bs) + ln(a+ bs)
b
(a+bs)
h′(s) = 2b ln(a+bs)(a+bs)
f ′(s) = g′(h(s))h′(s)
f ′(s) =
2b ln(a+ bs)
(a+ bs)
eln(a+bs) ln(a+bs)
f ′(s) =
2b ln(a+ bs)
(a+ bs)
eln(a+bs)
ln(a+bs)
f ′(s) =
2b ln(a+ bs)
(a+ bs)
(a+ bs)ln(a+bs)
v) f(x) = a
3x
b3x2−6x
f(x) =
g(x)
h(x)
f ′(x) =
g′(x)h(x)− g(x)h′(x)
h2(x)
g′(x) = (a3x)′ = 3a3x ln (a)
h′(x) = (b3x
2−6x) = (6x− 6)b3x2−6x ln (b)
4
f ′(x) =
3a3x ln (a)(b3x
2−6x)− a3x(6x− 6)(b3x2−6x) ln (b)
(b3x2−6x)2
f ′(x) =
3a3x ln (a)− a3x(6x− 6) ln (b)
(b3x2−6x)
f ′(x) =
a3x(3 ln (a)− (6x− 6) ln (b))
(b3x2−6x)
e)f(θ) = (θ2 + sec (θ) + 1)3
f ′(θ) = (θ2 + sec (θ) + 1)′3(θ2 + sec (θ) + 1)2
f ′(θ) = (2θ + sec (θ) tan (θ))3(θ2 + sec (θ) + 1)2
j)r(θ) =
√
2θsen(θ)
r′(θ) = (2θsen(θ))
1
2
(2θsen(θ))−
1
2
r′(θ) =
1
2
(2sen(θ) + 2θ cos (θ))(2θsen(θ))−
1
2
r′(θ) = (sen(θ) + θ cos (θ))(2θsen(θ))−
1
2
l)f(x) = 12x
2 cossec( 2x )
f(x) = g(x)h(x)
f ′(x) = g′(x)h(x) + g(x)h′(x)
g′(x) = (12x
2)′ = 122x = x
h′(x) = (cossec( 2x ))
′ = ( 2x )
′(−cossec( 2x )cotg(
2
x ))
h′(x) = −2x−2(−cossec( 2x )cotg(
2
x ))
f ′(x) = xcossec(
2
x
) +
1
2
x22x−2(cossec(
2
x
)cotg(
2
x
))
f ′(x) = cossec(
2
x
)(x+ cotg(
2
x
))
o)f(t) =
(
4t
t+1
)−2
5
f ′(t) =
(
4t
t+ 1
)′(
−2
(
4t
t+ 1
)−3)
(
4t
t+1
)′
= 4(t+1)−4t.1(t+1)2 =
4
(t+1)2
f ′(t) = −2 4
(t+ 1)2
(
4t
t+ 1
)−3
= −2 4(t+ 1)
3
(4t)3(t+ 1)2
f ′(t) = −8(t+ 1)
64t3
= − (t+ 1)
8t3
q)r(θ) =
(
sen(θ)
cos(θ)−1
)2
r′(θ) =
(
sen(θ)
cos(θ)− 1
)′
2
(
sen(θ)
cos(θ)− 1
)
(
sen(θ)
cos(θ)−1
)′
= cos(θ)(cos(θ)−1)−sen(θ)(−sen(θ))(cos(θ)−1)2(
sen(θ)
cos(θ)−1
)′
= cos
2(θ)−cos(θ)+sen2(θ)
(cos(θ)−1)2 = −
1−cos(θ)
(cos(θ)−1)2 =
1
(cos(θ)−1)
r′(θ) =
1
(cos(θ)− 1)
2
(
sen(θ)
cos(θ)− 1
)
=
2sen(θ)
(cos(θ)− 1)2
s)f(x) = x2e−
2
x
f ′(x) = 2x(e−
2
x ) + x2
2
x2
(e−
2
x )
f ′(x) = 2(e−
2
x )(x+ 1)
t)f(x) = ln (sen2(x))
f ′(x) = (sen2(x))′
1
sen2(x)
f ′(x) = 2cos(x)sen(x)
1
sen2(x)
f ′(x) =
2cos(x)
sen(x)
= 2cotg(x)
x) f(x) = log5(3x− 7)
(loga(y))
′ = 1y ln(a)
f ′(x) = (3x− 7)′ 1
(3x− 7) ln (5)
=
3
(3x− 7) ln (5)
6
z)f(x) = (1 + x2)earctg(x)
f ′(x) = 2xearctg(x) + (1 + x2)(arctg(x))′earctg(x)
y = (arctg(x))
tg(y) = x
d
dx (tg(y)) =
d
dx (x) = 1
(sec2 (y)) dydx = 1
dy
dx =
1
sec2(y) =
1
tg2(y)+1 =
1
tg2(arctg(x))+1 =
1
x2+1
f ′(x) = 2xearctg(x) + (1 + x2)
1
x2 + 1
earctg(x) = 2xearctg(x) + earctg(x)
f ′(x) = (2x+ 1)earctg(x)
7