Test #2
Math 1110-102
Name:
July 21st , 2014
No calculators.
Answer Key
Show your work!
1. (11 points) Implicit differentiation.
(a) Use implicit differentiation to find
(x, y) = (1, −2).
dy
given x2 + y 2 = 5. Then find the equation of the line tangent to x2 + y 2 = 5 at
dx
2x + 2yy 0 = 0
dy So the slope of the tangent at (x, y) = (1, −2) is
dx Differentiate and get:
=⇒
(x,y)=(1,−2)
2yy 0 = −2x
x =− y
y0 = −
=⇒
=−
(x,y)=(1,−2)
x
y
1
1
= . Therefore the tangent
−2
2
1
5
1
line’s equation is y − (−2) = (x − 1). Thus y = x − .
2
2
2
dy
given x2 y + y 3 + x4 = 22.
dx
When differentiating “x2 y” we need to use the product rule. The term “y 3 ” requires the chain rule.
(b) Use implicit differentiation to find
2xy + x2 y 0 + 3y 2 y 0 + 4x3 = 0
Differentiate and get:
(x2 + 3y 2 )y 0 = −(2xy + 4x3 )
=⇒
(2xy + 4x3 ) + (x2 + 3y 2 )y 0 = 0
=⇒
y0 = −
=⇒
2xy + 4x3
x2 + 3y 2
2. (12 points) Taylor Polynomials
(a) Find the third degree Taylor polynomial centered at x = 1 for f (x) = ln(x).
1
= x−1 , f 00 (x) = −x−2 , and f 000 (x) = 2x−3 . Next, we need to
x
evaluate the function and these derivatives at our base point: f (1) = ln(1) = 0, f 0 (1) = 1/1 = 1, f 00 (1) = −1/12 = −1,
f 000 (1) = 2/13 = 2. Finally, we plug these values into our Taylor polynomial formula. . .
We need the first 3 derivatives of f (x) = ln(x). f 0 (x) =
P3 (x) = 0 + 1(x − 1) +
2
1
−1
1
(x − 1)2 + (x − 1)3 = (x − 1) − (x − 1)2 + (x − 1)3
2
3!
2
3
(b) Suppose that g(x) is a function whose fifth degree Taylor polynomial centered at x = −2 is
P5 (x) = 5 + 3(x + 2) − 2(x + 2)2 + (x + 2)5
Either fill in the blank or circle “Not Enough Info.” if there isn’t enough information to find the answer.
First, recall that the linearization is just the first order Taylor polynomial. To get this we merely delete all of the terms
f 000 (−2)
f 00 (−5)
= −2,
= 0,
of orders 2 and above. Next, our polynomial tells us that: f (−2) = 5, f 0 (−2) = 3,
2
3!
(4)
(5)
f (−2)
f (−2)
= 0, and
= 1 (read off the coefficients of (x + 2)k for k = 0, 1, . . . , 5). Since we only have P5 (x) (and
4!
5!
not P6 (x)) there is no way to tell what f ( 6)(−2) might be – in fact, it might not even exist!
The linearization of g(x) based at x = −2 is
000
g (−2)
=
0
g
(6)
(−2)
=
5 + 3(x + 2)
g 0 (−2)
=
3
g 00 (−2)
=
−4
Not Enough Info.
3. (10 points) Use Newton’s method to approximate a solution of x3 = 2. Use the initial guess x0 = 1 and work through
2 iterations. Please simplify x1 . You may leave x2 unsimplified.
We need to solve f (x) = x3 −2. We have f 0 (x) = 3x2 so that our Newton’s iteration is xn+1 = xn −
4 3
−2
4
91
3
x0 = 1
x2 =
−
=
≈ 1.263888889.
4 2
3
72
3 3
√
Not too bad for 2 iterations. Maple’s approximation is: 3 2 ≈ 1.259921050.
13 − 2
−1
4
x1 = 1 −
=1−
=
3(12 )
3
3
1
x3n − 2
f (xn )
=
x
−
.
n
f 0 (xn )
3x2n
4. (12 points) Some Theory
(a) Looking at a plot of y = ln(x) and y = x − 1, you could see that the graph of natural log lies below the diagonal line
y = x − 1. Use the racetrack principal to justify why this is true for x ≥ 1. [Note: ln(1) = 0.]
Notice that ln(1) = 0 = 1 − 1. This means both y = ln(x) and y = x − 1 start at the same
value (i.e. 0) when x = 1.
d
1
d
Next, notice that
[ln(x)] =
≤ 1 =
[x − 1] as long as x ≥ 1. This means that
dx
x
dx
y = x − 1 grows faster than y = ln(x).
Therefore, by the Racetrack Principal, ln(x) ≤ x − 1 for all x ≥ 1.
(b) Suppose f (2) = 3 and f (5) = 12 and suppose that f is differentiable everywhere. Then the mean value theorem (MVT)
guarantees that there is some c such that. . .
f 0 (c) =
f (5) − f (2)
12 − 3
9
=
= =3
5−2
5−2
3
and
2
<
c
<
5
.
In other words, f 0 matches the average value of f on the interval [2, 5] somewhere inside that interval.
√
(c) Let g(x) = 3 x. Circle all that apply (their hypotheses are satisfied). MVT = Mean Value Theorem and IVT =
Intermediate Value Theorem.
g(x) on the interval [−1, 1]
MVT
/
IVT
√
1
g(x) = 3 x = x1/3 is continuous everywhere. However, g 0 (x) = x−2/3 is not defined at x = 0. So since g(x) is not
3
differentiable at x = 0, the MVT’s hypotheses are not satisfied. On the other hand, the IVT only assumes continuity, so
its hypotheses are satisfied.
5. (15 points) Derive the formula for the derivative.
(a) Find
d
[sec(x)] by writing secant in terms of cosine, differentiating, and simplifying.
dx
d
1
sin(x)
d
1
0 cos(x) − 1(− sin(x))
sin(x)
[sec(x)] =
=
or =
·
= sec(x) tan(x)
=
dx
dx cos(x)
cos2 (x)
cos2 (x)
cos(x) cos(x)
d
[arctan(x)] using the derivative of tan(x) and the chain rule. Specifically: (1) State the derivative of tangent,
dx
d
[arctan(x)], (3) Draw a triangle, and (4) Simplify your
(2) Write down an equation, differentiate it, and solve for
dx
trig/inverse trig expression for the derivative of arctangent.
(b) Find
d
[tan(x)] = sec2 (x) (this formula’s derivation is very similar to the derivative of the derivative of secant – whose
dx
derivative was found in part (a)). Next, tan(arctan(x)) = x (by definition, arctan and tangent are inverses). Differentiate
d
this equation and get: sec2 (arctan(x)) ·
[arctan(x)] = 1 (the left hand side follows from the chain rule).
dx
We need to simplify the trig/inverse trig expression appearing in our equation so we draw a
triangle. Our triangle needs to be drawn so that tan(θ)√= x (i.e. arctan(x) = θ). So we
should make x the opposite side, 1 the adjacent side, and 1 + x2 the hypotenuse (this follows
from the Pythagorean theorem).
Therefore, since secant is hypotenuse over adjacent, we get
√
sec2 (arctan(x)) = sec2 (θ) = ( x2 + 1/1)2 = x2 + 1.
1
1
d
Therefore,
[arctan(x)] =
= 2
.
dx
sec2 (arctan(x))
x +1
First,
2
6. (20 points) Derivatives. Please simplify your answers.
(a) h(1) = 2, h0 (1) = 3, g(0) = 1, g 0 (0) = 5. Let m(x) = h(g(x)). Compute m0 (0).
We just need the chain rule: m0 (x) = h0 (g(x)) · g 0 (x). Therefore, m0 (0) = h0 (g(0)) · g 0 (0) = h0 (1) · g 0 (0) = 3 · 5 = 15 .
(b) Let y = sin(2x) + arcsin(x) + cosh(5x + 1) + ln |x| +
y = sin(2x)+arcsin(x)+cosh(5x+1)+ln |x|+x−1/2
(c) Let y = x5 cos(3x). Find
√1 .
x
Find y 0 .
y 0 = cos(2x) · 2 + √
=⇒
1 1
1
+ sinh(5x + 1) · 5 + − x−3/2
x 2
1 − x2
d2 y
.
dx2
Use the product rule and then again: y 0 = 5x4 cos(3x) + x5 (− sin(3x)) · 3 = 5x4 cos(3x) − 3x5 sin(3x).
y 00 = 20x3 cos(3x) + 5x4 (− sin(3x) · 3) − 15x4 sin(3x) − 3x5 cos(3x) · 3 = 20x3 cos(3x) − 15x4 sin(3x) − 15x4 sin(3x) −
9x5 cos(3x) = (20x3 − 9x5 ) cos(3x) − 30x4 sin(3x) .
(d) Let f (x) = ln
!
√
(x + 4)7 x3 + 1
. Find f 0 (x) and simplify Hint: Use laws of logs. Don’t do it the “hard way”.
x100 e−2x
f (x) = ln (x + 4)7 (x3 + 1)1/2 − ln x100 e−2x = ln (x + 4)7 + ln (x3 + 1)1/2 − ln(x100 ) − ln(e−2x )
= 7 ln(x + 4) +
1
ln(x3 + 1) − 100 ln(x) + 2x
2
=⇒
f 0 (x) =
1
3x2
100
7
+ · 3
−
+2
x+4 2 x +1
x
7. (20 points) More derivatives. You do not need to simplify your answers here.
(a) Let y = (x5 + 3x2 + 1)50 . Find y 0 .
Chain rule
y 0 = 50(x5 + 3x2 + 1)49 · (5x4 + 6x)
(b) Let y =
e−x + 1
. Find y 0 .
ln(x) + x2 + 5
Quotient rule
(−e−x )(ln(x) + x2 + 5) − (e−x + 1)
y =
(ln(x) + x2 + 5)2
0
2
(c) Let y = ln(sin(ex )). Find y 0 .
Multiple instances of the chain rule
y0 =
(d) Let y = 4arcsin(x) . Find y 0 .
2
2
1
· cos(ex ) · ex · 2x
2
x
sin(e )
Chain rule
y 0 = 4arcsin(x) ln(4) · √
3
1
1 − x2
1
x
+ 2x