Math 134 Fall 2011 Quiz 4
October 17, 2011
Name:
1. Find the equation for the tangent line to the curve y = 3ex (2x + cos(x)) when x = 0.
First we begin by finding the y value when x = 0 which is given by y = 3e0 (2(0)+cos(0)) = 3.
Thus the tangent line passes through the point (0, 3). Next we find the slope at this point
by evaluating the derivative and then plugging in 0.
d
d
[3ex ] (2x + cos(x)) + 3ex [2x + cos(x)]
dx
dx
= 3ex (2x + cos(x)) + 3ex (2 − sin(x))
f 0 (x) =
If we evaluate at x = 0 we get:
f 0 (0) = e0 (2(0) + cos(0)) + 3e0 (2 − sin(0)) = 1 + 3(2) = 7
Thus we can find the equation of the tangent line by using (0, 3) and a slope of 7. The
equation is then:
(y − 3) = 7(x − 0)
y = 7x + 3
2. Below is a table of values for g(x) and g 0 (x) for several different values of x.
x g(x) g 0 (x)
0
1
2
0.5 1.5
-1
1
0
1.5
1.5
1
-1
2
.5
-7
Given that l(x) = g(g(x)). Find l0 (2).
By the chain rule l0 (x) = g 0 (g(x))g 0 (x), so l0 (2) = g 0 (g(2))g 0 (2) this simplifies to l0 (2) =
g 0 (.5)(−7) = (1.5)(7) = 10.5.
1
3. Compute the derivative of the following functions, you do not need to simplify your
answers.
(a)
f (x) = e(4x+cos(3x)) .
g(x) = eg(x) g 0 (x) thus I have that:
e
d
0
(cos(3x)) d
(cos(3x))
f (x) = e
4 − sin(3x) [3x]
[4x + cos(3x)] = e
dx
dx
Using the chain rule I remember that
d
dx
= e(cos(3x)) (4 − sin(3x)3) = (4 − 3 sin(3x)e(cos(3x))
Notice that this problem involves using the chain rule twice.
(b)
√
2 + x4
g(x) =
.
sin(πx)
First I use the quotient rule to write:
√
√
d
d
4 −
2
+
x
2 + x4 dx
[sin(πx)]
sin(πx)
0
dx
g (x) =
2
(sin(πx))
√
−1/2
−1/2
(4x3 ) = 2x3 (2 + x4 )
using the chain
The derivative of 2 + x4 = 12 (2 + x4 )
rule. Also the derivative of sin(πx) is equal to π cos(πx) using the chain rule. Thus
we have:
√
2x3 (2 + x4 )−1/2 sin(πx) − π 2 + x4 cos(πx)
0
g (x) =
(sin(πx))2
2