Powers, Exponentials, Products, and Quotients
Spring 2014
BC1
Name: ________________________
1) Differentiate each of the following. Do not simplify.
1
(1 pt. each) a) 3x 4 − 4x 2/3 +
b) 4⋅5x
x
Sum, constant multiple, and
power rules:
8
1
12x 3 − x −1/2 − x −3/2
3
2
(2 pts. each)
b) 2x⋅(5x3 – 7)
Product rule:
2x⋅ln(2)⋅(5x3 – 7) + 2x⋅(15x2)
(3 pts. each)
e)
cos(x)
x ⋅ ex
Quotient rule, with product rule
to figure derivative of
denominator:
xe x (− sin(x)) − (e x + xe x )cos(x)
(xe x )2
Constant multiple and
exponential rules:
4⋅5x⋅ln(5)
d)
x 3 + 3x + 2
ex + x
Quotient rule:
(e x + x )(3x 2 + 3) − (e x + 1 / 2 x )(x 3 + 3x + 2)
(e x + x )2
f) xe x sin(x) x + 1
Product rule used over and over!
e x sin(x) x + 1
+xe x sin(x) x + 1
+xe x cos(x) x + 1
+xe x sin(x)
1
2 x +1
2) (5 pts.) Shown at right is the graph of
y = 16 − x . Consider a rectangle with one
corner at the origin, one on the positive x-axis,
one on the positive y-axis, and one on the curve.
Let the width of the rectangle be x. Find x that
makes the area of this rectangle a maximum.
Justify your answer with calculus explanations.
(You should probably draw the rectangle to help
understand what is going on)
4
3
2
16 - x
1
x
5
10
15
From the picture, we see that the rectangle has an area of A(x) = x 16 − x . To maximize, we
will find where the derivative crosses from positive to negative. Using the product rule to
x
differentiate, we get A′(x) = 16 − x −
. We need this to equal zero, so:
2 16 − x
x
16 − x −
=0
2 16 − x
x
16 − x =
2 16 − x
2(16 − x) = x
32 − 2x = x
x = 32 / 3
If we put in smaller x (like x = 7, to make the square root convenient) we get positive derivatives,
and if we put in larger x (like x = 12 for a convenient square root) the derivative is negative. So
since the derivative goes from positive to negative at x = 32/3 we have found a maximum for the
function.
3) Let f(x) be a function whose derivative is
graphed at right. The derivative goes through the
point at (1, 4), and f(1) = 6.
Let g(x) =
5 − x and let h(x) = f(x)⋅g(x).
a) (2 pts.) Explain why f(2) must be positive.
f(1) is positive, and f′(x) is positive so f(x) is
increasing. So f(2) > f(1), so is positive.
4
3
2
1
0.5
1.0
1.5
2.0
2.5
3.0
b) (2 pts.) Determine h′(1).
h′(x) = f′(x)g(x) + f(x)g′(x) by the product rule. We know f(1) = 6, f′(1) = 4, g(1) = 4 = 2 and
−1
g′(x) =
so g′(1) = –1/4. Putting these together tells us h′(x) = 4⋅2 – 6/4 = 13/2.
2 5− x
c) (3 pts.) Determine whether the graph of h(x) is concave up or concave down at x = 2.
Concavity involves h′′, so we differentiate the expression for h′ to get h′′ = f’’g + 2f′g′ + fg′′.
Now plugging in x = 2, we don’t actually know f′′(2), but from the graph of f′ decreasing at x = 2
we know f is concave down, so f′′(2) < 0. We also know f′(2) is positive by looking at the graph,
and f(2) is positive from part a). We can calculate g(2) is positive, g′(2) is negative, and we can
−1
calculate g′′(x) = (5 − x)3/2 which is also negative at x = 2. So we find that h′′(2) is
4
(–⋅+) + 2(+⋅–) + (+⋅–) = negative, so h is concave down at x = 2.