Workshop #15
Professor D. Olles
1. Find
dy
dx
by implicit differentiation.
3
a. x + y 3 = 1
b. x2 + xy − y 2 = 4
c. 4 cos x sin y = 1
d. ex/y = x − y
e. tan (x − y) =
y
1+x2
2. Use implicit differentiation to find an equation for the tangent line to the
curve at the given point.
x2 + 2xy − y 2 + x = 2, (1, 2)
3. Find y 00 by implicit differentiation.
9x2 + y 2 = 9
4. Use implicit differentiation to derive the formula for the derivative of y =
loga x.
5. If f (x) = sin−1 x, show that f 0 (x) =
√ 1
.
1−x2
6. If f (x) = tan−1 x, show that f 0 (x) =
1
1+x2 .
2
7. Find an equation of the line tangent to the curve y = ln (xex ) at (1,1).
8. Use logarithmic differentiation to fin the derivative of the function.
y=
√
2
xex (x2 + 1)10
9. Find y 0 if xy = y x .
10. Find a formula for f (n) (x) if f (x) = ln (x − 1).
1
Solutions
1. Find
dy
dx
by implicit differentiation.
a. x3 + y 3 = 1
3x2 + 3y 2
dy
=0
dx
dy
= −3x2
dx
dy
x2
=− 2
dx
y
3y 2
b. x2 + xy − y 2 = 4
2x + (1)y + x
dy
dy
− 2y
=0
dx
dx
dy
dy
−x
dx
dx
dy
2x + y = (2y − x)
dx
2x + y
dy
=
dx
2y − x
2x + y = 2y
c. 4 cos x sin y = 1
dy
4 − sin x sin y + cos x cos y
=0
dx
−4 sin x sin y + 4 cos x cos y
dy
=0
dx
dy
= 4 sin x sin y
dx
dy
sin x sin y
=
= tan x tan y
dx
cos x cos y
4 cos x cos y
d. ex/y = x − y
ex/y
e
x/y
dy
y(1) − x dx
y2
1
x dy
− 2
y y dx
!
=1−
dy
dx
=1−
dy
dx
xex/y dy
dy
ex/y
−
=1−
y
y 2 dx
dx
2
xex/y dy
dy
ex/y
−1=
−
y
y 2 dx dx
x/y
ex/y
dy
xe
−
1
−1=
y
y2
dx
x/y
ex/y − y
xe
− y 2 dy
=
y
y2
dx
x/y
dy
e
−y
y2
=
x/y
y
dx
xe
− y2
dy
yex/y − y 2
= x/y
dx
xe
− y2
e. tan (x − y) =
y
1+x2
dy
(1 + x2 ) dx
− y(2x)
2
(1 + x )2
1
dy
dy
2xy
2
2
sec (x − y) − sec (x − y)
=
−
dx
1 + x2 dx (1 + x2 )2
2xy
1
dy
dy
sec2 (x − y) +
=
+ sec2 (x − y)
(1 + x2 )2
1 + x2 dx
dx
1
dy
2xy
2
=
+
sec
(x
−
y)
sec2 (x − y) +
2
2
2
(1 + x )
1+x
dx
2
2
2 2
2
1 + (1 + x ) sec (x − y) dy
(1 + x ) sec (x − y) + 2xy
=
(1 + x2 )2
1 + x2
dx
2 2
2
2
(1 + x ) sec (x − y) + 2xy
dy
1+x
=
(1 + x2 )2
1 + (1 + x2 ) sec2 (x − y)
dx
dy
sec (x − y) 1 −
dx
2
=
dy
(1 + x2 )2 sec2 (x − y) + 2xy
=
dx
1 + x2 + (1 + x2 )2 sec2 (x − y)
2. Use implicit differentiation to find an equation for the tangent line to the
curve at the given point.
x2 + 2xy − y 2 + x = 2, (1, 2)
dy
dy
− 2y
+1=0
dx
dx
dy
2x + 2y + 1 = (2y − 2x)
dx
2x + 2y + 2x
3
dy
2x + 2y + 1
=
dx
2y − 2x
mt =
dy
2+4+1
7
=
(1,2) =
dx
4−2
2
y − y1 = mt (x − x1 )
7
(x − 1)
2
7
7
y−2= x−
2
2
7
3
y = x−
2
2
y−2=
3. Find y 00 by implicit differentiation.
9x2 + y 2 = 9
18x + 2yy 0 = 0
2yy 0 = −18x
y0 = −
18x
2y
y0 = −
y 00 = −
y 00 = −
9x
y
(y)(9) − (9x)(y 0 )
y2
(y)(9) − (9x)(− 9x
y )
y2
00
y =−
y 00 = −
81x2
y
y2
9y +
9y 2 + 81x2
y3
4. Use implicit differentiation to derive the formula for the derivative of y =
dy
loga x. Let y = loga x. We need to find the derivative, dx
.
y = loga x
ay = x
d y
d
(a ) =
(x)
dx
dx
4
dy
=1
dx
dy
1
= y
dx
a ln a
1
dy
= log x
a
dx
a
ln a
dy
1
=
dx
x ln a
ay ln a
5. If f (x) = sin−1 x, show that f 0 (x) =
√ 1
.
1−x2
f (x) = sin−1 x
y = sin−1 x
sin y = x
d
d
(sin y) =
(x)
dx
dx
dy
=1
cos y
dx
dy
1
=
dx
cos y
From the claim that sin y = x, we may set up the triangle:
√
Which gives us cos y =
1−x2
.
1
So,
f 0 (x) = √
6. If f (x) = tan−1 x, show that f 0 (x) =
1
1 − x2
1
1+x2 .
f (x) = tan−1 x
y = tan−1 x
tan y = x
5
dy
=1
dx
dy
1
=
dx
sec2 y
sec2 y
From the claim that tan y = x, we may set up the triangle:
Which gives us sec y =
√
1 + x2 . So,
dy
1
= √
dx
( 1 + x2 )2
f 0 (x) =
1
1 + x2
2
7. Find an equation of the line tangent to the curve y = ln (xex ) at (1,1).
ex2 (1 + 2x2 )
2
2
dy
1 1 + 2x2
= x2 (1)ex + x(ex (2x)) =
=
2
dx
x
xe
xex
dy
1+2
=3
x=1 =
dx
1
y − y1 = mt (x − x1 )
mt =
y − 1 = 3(x − 1)
y − 1 = 3x − 3
y = 3x − 2
8. Use logarithmic differentiation to fin the derivative of the function.
√
2
xex (x2 + 1)10
h√ 2
i
ln y = ln xex (x2 + 1)10
h√ 2 i
ln y = ln xex + ln (x2 + 1)10
h
i
h 2i
ln y = ln x1/2 + ln ex + ln (x2 + 1)10
y=
6
1
ln x + x2 ln e + 10 ln (x2 + 1)
2
1
ln y = ln x + x2 + 10 ln (x2 + 1)
2
11
1
1 dy
=
+ 2x + 10 2
(2x)
y dx
2x
x +1
ln y =
1
20x
1 dy
=
+ 2x + 2
y dx
2x
x +1
y
20xy
dy
=
+ 2xy + 2
dx
2x
x +1
dy
=
dx
√
√ 2
2
√ 2
20x xex (x2 + 1)10
xex (x2 + 1)10
+ 2x xex (x2 + 1)10 +
2x
x2 + 1
9. Find y 0 if xy = y x .
ln xy = ln y x
y ln x = x ln y
d
d
[y ln x] =
[x ln y]
dx
dx
dy
1
1 dy
ln x + y = (1) ln y + x
dx
x
y dx
ln x
y
x dy
dy
+ = ln y +
dx x
y dx
dy
x dy
y
−
= ln y −
dx y dx
x
x dy
y
ln x −
= ln y −
y dx
x
ln x
ln y − xy
dy
=
dx
ln x − xy
x ln y−y
x
dy
=
y ln x−x
dx
y
dy
xy ln y − y 2
=
dx
xy ln x − x2
7
10. Find a formula for f (n) (x) if f (x) = ln (x − 1).
f (x) = ln (x − 1)
1
= (x − 1)−1
x−1
1
f 00 (x) = −(x − 1)−2 = −
(x − 1)2
f 0 (x) =
f 000 (x) = 2(x − 1)−3 =
2
(x − 1)3
f (4) (x) = −6(x − 1)−4 = −
f (5) (x) = 24(x − 1)−5 =
6
(x − 1)4
24
(x − 1)5
If we allow the first derivative to represent the case where n = 1, the
second n = 2 and so on, then we can find a formula in terms of n for the
nth derivative.
The alternating signs give us negative functions when n is even, and positive functions when n is odd. So, a factor that should exist in the numerator is (−1)n+1 .
The power of the binomial in the denominator is always equal to the value
for n in that function, so we need (x − 1)n in the denominator.
The value in the numerator is increasing as a factorial. But, when n = 3,
the numerator is 2 and 3! 6= 2. So, we need (n − 1)! in the numerator.
f (n) (x) =
(−1)n+1 (n − 1)!
(x − 1)n
8