Math 170 Calculus I
Exam III Review Solutions
1. Find dy/dx, using any method you prefer.
(a) y = ln(x3 + 4x2 − 3x + 5)
3x2 + 8x − 3
y0 = 3
x + 4x2 − 3x + 5
(b) y = ln(ex )
y = ln(ex ) = x, so y 0 = 1
p
(c) y = ln(2x3 )
3
6x
p
y 0 = 12 (ln(2x3 ))−1/2 3 =
2
2x
2x ln(2x3 )
ex
ln(x)
ln(x)ex − ex x1
y0 =
(ln(x))2
(d) y =
(e) y = ex sin−1 (x)
1
y 0 = ex sin−1 (x) + ex √
1 − x2
−1
3
(f) y = tan (4x )
1
12x2
2) =
y0 =
(12x
1 + (4x3 )2
1 + 16x6
!
1 + ex
(g) y = ln
1 − ex
#
"
(1 − ex )(ex ) − (1 + ex )(−ex )
2ex
d 1 + ex
=
=
dx 1 − ex
(1 − ex )2
(1 − ex )2
2ex
2ex (1 − ex )
2ex
(1 − ex )2
y0 =
=
=
(1 − ex )2 (1 + ex ) (1 − ex )(1 + ex )
1 + ex
1 − ex
√
2
(h) y = e √1−3x
√
2
2
y 0 = e 1−3x 21 (1 − 3x2 )−1/2 (−6x) = −3x(1 − 3x2 )−1/2 e 1−3x
(i) y = sin−1 (x) + cos−1 (x)
1
1
y0 = √
−√
=0
1 − x2
1 − x2
2. Find y 0 :
(a) x2 + y 2 = 5
2x + 2yy 0 = 0
2yy 0 = −2x
− 2x
x
y0 =
=−
2y
y
2
3
(b) 3xy + 6y − 4x = 10
3y 2 + 6xyy 0 + 6y 0 − 12x2 = 0
6xyy 0 + 6y 0 = 12x2 − 3y 2
y 0 (6xy + 6) = 12x2 − 3y 2
12x2 − 3y 2 4x2 − y 2
y0 =
=
6xy + 6
2xy + 2
(c) tan(xy) = y
sec2 (xy)(y + xy 0 ) = y 0
sec2 (xy)y + sec2 (xy)xy 0 = y 0
sec2 (xy)y = y 0 − sec2 (xy)xy 0 = y 0 (1 − sec2 (xy)x)
y sec2 (xy)
0
y =
1 − x sec2 (xy)
3. Find y 00 if x2 + y 2 = 50. Simplify your answer as much as possible.
2x + 2yy 0 = 0
x
y0 = −
y
y − xy 0
y − x(−x/y)
y 2 + x2
y 00 = −
=
−
=
−
y2
y2
y3
4. Find y 0 using logarithmic differentiation if
!
x2 − 5
(a) y =
(x3 + 1)(3x4 − 5x2 + 2x)
3x4 + 6
ln(y) = ln(x2 − 5) − ln(3x4 + 6) + ln(x3 + 1) + ln(3x4 − 5x2 + 2x)
2x
12x3
3x2
12x3 − 10x + 2
y0
=
−
+
+
y
x2 − 5 3x4 + 6 x3 + 1 3x4 − 5x2 + 2x !
!
3
2
3 − 10x + 2
2−5
2x
12x
3x
12x
x
y0 =
−
+
+
(x3 + 1)(3x4 − 5x2 + 2x)
x2 − 5 3x4 + 6 x3 + 1 3x4 − 5x2 + 2x
3x4 + 6
!4
√
x(x2 − 3)
(b) y = √
3
5x3 − 4x + 1
√
√
ln(y) = 4 ln( x(x2 − 3)) − 4 ln( 3 5x3 − 4x + 1) = 2 ln(x) + 4 ln(x2 − 3) − 43 ln(5x3 − 4x + 1)
8x
4(15x2 − 4)
2
y0
+
−
=
y
x x2 − 3 3(5x3 − 4x + 1)
!
!4
√
2 − 4)
2 − 3)
2
8x
4(15x
x(x
√
y0 =
+
−
3
x x2 − 3 3(5x3 − 4x + 1)
5x3 − 4x + 1
5. Two parallel sides of a rectangle are being lengthened at the rate of 2 in/sec, while the other two sides are
shortened in such a way that the figure remains a rectangle with constant area 50 in2 . What is the rate
of change of the perimeter of the rectangle when the length of an increasing side is 5 in? Is the perimeter
increasing or decreasing?
Let x be the length of a side that’s getting longer, and y be the length of a side that’s getting shorter.
Note that we’re given dx/dt = 2. The area and perimeter of the rectangle would be
50 = xy,
P = 2x + 2y
The rate of change of the perimeter would be (taking derivatives with respect to x):
dP
dx
dy
dy
=2
+2
= 2(2) + 2
dt
dt
dt
dt
We need to go back and find dy/dt, using the area formula’s derivative:
d
dy
dx
[50] = x + y
dt
dt
dt
dy
+ 2y
dt
When x = 5, we must have y = 10, since the area is fixed at 50. Now
0=x
0=5
dy
+ 2(10)
dt
so dy/dt = −4. Then
dP
dy
=4+2
= 4 + 2(−4) = −4
dt
dt
So the perimeter is decreasing at a rate of 4 in/sec.
6. Find the limits:
(a)
x60
x→+∞ ex
lim
This is type ∞/∞, so we can use L’Hopital’s rule:
x60
60x59
=
lim
lim
x→+∞ ex
x→+∞ ex
That didn’t help! But we can keep doing it. Eventually, after taking enough derivatives, the numerator
will be 0. The denominator never changes, since it’s derivative is always just ex . So after taking 61
derivatives, we get
0
x60
= lim x = 0
lim
x→+∞ e
x→+∞ ex
(b)
5x3 − 4x + 3
x→+∞
2x2 − 1
This is also type ∞/∞, so we can use L’Hopital’s rule (possibly more than once!):
lim
5x3 − 4x + 3
15x2 − 4
30x
=
lim
= lim
=∞
2
x→+∞
x→+∞
x→+∞ 4
2x − 1
4x
lim
!
1
1
(c) lim
−
x→0 x
sin(x)
This one is type ∞ − ∞. We can’t use L’Hopital’s rule quite yet. We first need to rewrite the
expression as something of type 0/0 or ∞/∞. The easiest way (in this case) is to find a common
denominator:
!
1
1
sin(x) − x
lim
−
= lim
x→0 x
x→0
sin(x)
x sin(x)
Now we can repeatedly use L’Hopital’s rule, stopping whenever we get a limit we can evaluate:
cos(x) − 1
sin(x) − x
lim
= lim
x→0
x→0 x cos(x) + sin(x)
x sin(x)
− sin(x)
= lim
=0
x→0 x sin(x) + cos(x) + cos(x)
(d)
√
lim ( x2 + x − x)
x→+∞
Multiply by something equivalent to 1: a fraction with the same numerator and denominator, where
both are the conjugate of the expression you started with:
√
p
p
x2 + x + x
x
2
2
lim ( x + x − x) = lim ( x + x − x) √
= lim √
2
2
x→+∞
x→+∞
x→+∞
x +x+x
x +x+x
This is an indeterminate form (which type?) so we can use L’Hopital’s rule:
p
x
lim ( x2 + x − x) = lim √
2
x→+∞
x→+∞
x +x+x
1
= lim 1 2
−1/2
x→+∞ (x + x)
(2x) + 1
2
1
1
1
= lim
=
= lim p
x→+∞ x(x2 + x)−1/2 + 1
x→+∞
2
1 + 1/x + 1
(e) lim (e2x − 1)x
x→0+
NOT ON THE EXAM