Math 135 Test 2 Sample Solutions March 28, 2014

Math 135

Test 2 Sample Solutions March 28, 2014

Show your algebra whenever calculations are done by hand.

1. (20 pts.) Shorter answers.

(a) Compute the derivative of f (x) = √

1 + x³ using the limit definition of the deriva- tive.

Start with the difference quotient for f (x):

f (x + h) − f (x)

h = p1 + (x + h)³−√ 1 + x³

h .

To simplify this expression, multiply by the conjugate in the numerator and denominator and simplify:

p1 + (x + h)³−√ 1 + x³ h

p1 + (x + h)³+√ 1 + x³ p1 + (x + h)³+√

1 + x³ = (1 + (x + h)³) − (1 + x³) h(p1 + (x + h)³+√

1 + x³) Now we have to expand the terms in the numerator and cancel terms

where possible:

(3x²h + 3xh² + h³ h(p1 + (x + h)³+√

1 + x³) = 3x²+ 3xh + h² p1 + (x + h)³+√

1 + x³. We use this in the limit:

f⁰(x) = lim

h→0

f (x + h) − f (x) h

= lim

h→0

3x²+ 3xh + h² p1 + (x + h)³+√

1 + x³

= 3x²

√1 + x³ +√ 1 + x³

= 3 2

x²

√1 + x³

(b) Sketch the graph of the inverse function for y = tan(x) where the domain of the tangent is taken to be (^π₂,^3π₂ ). (Hint: What is the range of the inverse?)

Figure 1:

Figure 2: The graph of y = f (x) is shown in red.

2. (20 pts.) Figure 2 shows the graph of a function y = f (x).

(a) On a separate set of axes, sketch the graph of the derivative of f .

Figure 3: The graph of y = f⁰(x) is shown in red. Notice the vertical asymptotes in green.

(b) Where are the discontinuities of f⁰ located?

The discontinuities are located at x = 0.4, x = 1.5, and x = 2.5. The derivative is not defined at these points.

(c) What are the types of the discontinuities of f⁰?

The discontinuities at x = 0.5 and x = 1.5 are infinite discontinuities and the discontinuity at x = 2.5 is a jump discontinuity.

3. (20 pts.) Evaluate the following derivatives:

(a) Q⁰(r) for Q(r) = sin(r) sin⁻¹(r).

Start with the product rule:

Q⁰(r) = cos(r) sin⁻¹(r) + sin(r) 1

√1 − r². (b) h⁰(b) for h(b) =p1 + (ln(b))^{3 2}.

Rewrite the cube root as the ¹₃ power and start with the power rule:

h⁰(b) = 1

3(1 + (ln(b))²)^−2/32 ln(b)1 b = 2

3b

ln(b) (1 + (ln(b))²)^2/3. (c) K⁰⁰⁰ for K(z) = 17^z/2.

Use the exponential rule three times:

K⁰(z) = ln(17) 2 17^z/2 K⁰⁰(z) = (ln(17)

2 )²17^z/2 K⁰⁰⁰(z) = (ln(17)

2 )³17^z/2

(d) _dy^d22 csc³(7y).

Start with the power rule, then use the product rule:

d

dycsc³(7y) = 3 csc²(7y)(− csc(7y) cot(7y))7 = −21 csc³(7y) cot(7y) d²

dy² csc³(7y) = 147 3 csc³(7y) cot²(7y) + csc⁵(7y)
4. (20 pts.) Define a function g(x) = _1+x^x2.

(a) Calculate g⁰(x). Find the intervals on which g is increasing and the intervals on which g is decreasing.

Find the intervals on which g⁰ is positive and on which it is negative.

First,

g⁰(x) = (1 + x²) − 2x²

(1 + x²)² = 1 − x² (1 + x²)².

The denominator of ⁰ is always positive, so the sign of g⁰ is determined by the sign of the numerator. The numerator is positive for x² < 1 and negative for x² > 1. Putting these together, g⁰ is positive and g is increasing on the interval (−1, 1); g⁰ is negative and g is decreasing on the intervals (−∞, −1) and (1, ∞).

(b) Calculate g⁰⁰(x). Find the intervals on which g⁰ is increasing and the intervals on which g⁰ is decreasing.

First calculate g⁰⁰(x):

g⁰⁰(x) = (−2x)(1 + x²)²− (1 − x²)2(1 + x²)2x (1 + x²)⁴

= (−2x)(1 + x²) − (1 − x²)4x (1 + x²)³

= 2x(x² − 3) (1 + x²)³ .

As above, g⁰will be increasing when g⁰⁰is positive and g⁰ will be decreasing when g⁰⁰ is negative. Again the denominator is positive so the sign of g⁰⁰ is determined by the sign of the numerator. The numerator 2x(x²− 3) will be positive when x and x²− 3 have the same sign and negative when they have opposite signs. Since x²− 3 is positive for x >√

3 and x < √ 3, we conclude that the numerator is positive when x >√

3. Similarly, the numerator will be positive if −√

3 < x < 0. The numerator will be negative on the remaining intervals, x < √

3 and 0 < x < √

3. Putting this together, g⁰ will be increasing on the intervals (−√

3, 0) and (√ 3, ∞);

g⁰ will be decreasing on the intervals (0,√

3) and (−∞, −√ 3).

(c) Are there any intervals on which g increasing and g⁰ is increasing? What are they?

Here we need to intersect the intervals we have in (a) and (b). First, g is increasing on the interval (−1, 1) and g⁰ will be increasing on the intervals (−√

3, 0) and (√

3, ∞). Intersecting these, we see that g and g⁰ are increasing on the interval (−1, 0).

5. (20 pts.) Figure 2 show the plot of a curve C defined implicitly by (x − y)²+ (2x + y)² = 4

Figure 4: The tangent line constructed in 5(c).

(a) Find a formula for ^dy_dx by implicit differentiation.

Differentiate both sides of the equation with respect to x. Notice we use the power rule twice on the left side along with the chain rule.

2(x − y)(1 − dy

dx) + 2(2x + y)(2 +dy

dx) = 0 (x − y)(1 − dy

dx) + (2x + y)(2 + dy

dx) = 0 x − y − (x − y)dy

dx + 2(2x + y) + (2x + y)dy

dx = 0 (5x + y) + (x + 2y)dy

dx = 0 (x + 2y)dy

dx = −(5x + y) dy

dx = −5x + y x + 2y

(b) Using algebra, find the coordinates of the point in the first quadrant where the curve has slope -1.

If −1 = ^dy_dx, then 1 = ^5x+y_x+2y and x + 2y = 5x + y or y = 4x. We’re not done yet because the point must lie on C. Substitute y = 4x into the formula for C and solve for x:

(x − 4x)²+ (2x + 4x)² = 4

=⇒ 9x²+ 36x² = 4

=⇒ 45x² = 4

=⇒ x = r 4

45 = 2 3

√5 5 . (c) Find the equation for the tangent line to the curve at that point.

Since x = ²₃

√ 5

5 and y = 4x, the y-coordinate of this point is y = ⁸₃

√ 5 5 . The slope is −1, so the equation for the line is

y = −(x −2 3

√5 5 ) + 8

3

√5 5 .