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Exam 2 Review Solutions
- True or False, and explain:
(a) The derivative of a polynomial is a polynomial. True. A polynomial is a function of the form a 0 +a 1 x+a 2 x^2 +... anxn, and its derivative will also have integer powers of x by the Power Rule, nxn−^1. (b) If f is differentiable, then (^) dxd
f (x) = f^ ′(x) 2
f (x) True: d dx
f (x) = d dx (f (x))^1 /^2 =
(f (x))−^1 /^2 f ′(x) = f ′(x) 2
f (x)
(c) The derivative of y = sec−^1 (x) is the derivative of y = cos(x). False. The notation, sec−^1 (x) is for the inverse secant function, which is not the reciprocal of the secant. For extra practice, to get the formula for the derivative of y = sec−^1 (x): sec(y) = x From this, draw a right triangle with one acute angle labelled y, the hypotenuse x and the adjacent length x. This gives the length of the side opposite:
x^2 − 1. Now differentiate:
sec(y) tan(y) dy dx
From the triangle, sec(y) = x and tan(y) =
x^2 − 1, so: dy dx
x
x^2 − 1
(d) (^) dxd (10x) = x 10 x−^1 False. The Power Rule can only be used for xn, not ax. The derivative is 10x^ ln(10). (e) If y = ln |x|, then y′^ = (^1) x. TRUE. To see this, re-write the function:
y =
ln(x) if x > 0 ln(−x) if x < 0 ⇒^ y
1 x^ if^ x >^0 −x ·^ (−1)^ if^ x <^0 From which we see that y′^ = (^1) x , x 6 = 0. (f) The equation of the tangent line to y = x^2 at (1, 1) is:
y − 1 = 2x(x − 1)
False. This is the equation of a parabola, not a line. The derivative, y′^ = 2x gives a formula for the slope of the tangent line, and is not the slope itself. To get the slope, we evaluate the derivative at x = 1, which gives y′^ = 2. The slope of the tangent line is therefore y − 1 = 2(x − 1). (g) If y = e^2 , then y′^ = 2e False. e^2 is a constant, so the derivative is zero. (h) If y = |x^2 + x|, then y′^ = | 2 x + 1|. False. First, rewrite y, then differentiate:
y =
x^2 + x if x ≤ 0 or x ≥ 1 −(x^2 + x) if 0 < x < 1 ⇒^ y
2 x + 1 if x < 0 or x > 1 −(2x + 1) if 0 < x < 1
Compare this to | 2 x + 1|:
| 2 x + 1| =
2 x + 1 if x ≥ − 1 / 2 −(2x + 1) if x < − 1 / 2
By the way, we also note that y is NOT differentiable at x = 0 or at x = 1 by checking to see what the derivatives are approaching as x → 0 and as x → 1. (i) If y = ax + b, then dyda = x True. dyda means that we treat a as an independent variable, and x, b as constants.
- Find the equation of the tangent line to x^3 + y^3 = 3xy at the point ( 32 , 32 ).
We need to find the slope, dydx
∣x=3/ 2 ,y=3/ 2
3 x^2 + 3y^2 y′^ = 3y + 3xy′^ ⇒ y′(3y^2 − 3 x) = 3y − 3 x^2 ⇒ y′^ = y − x^2 y^2 − x Substituting x = 3/ 2 , y = 3/2 gives y′^ = −1, so the equation of the tangent line is y − 3 /2 = −1(x − 3 /2)
- If f (0) = 0, and f ′(0) = 2, find the derivative of f (f (f (f (x)))) at x = 0.
A cute chain rule problem! Here we go- the derivative is:
f ′(f (f (f (x)))) · f ′(f (f (x))) · f ′(f (x)) · f ′(x)
Now subsitute x = 0 and evaluate:
f ′(f (f (f (0)))) · f ′(f (f (0))) · f ′(f (0)) · f ′(0) f ′(0) · f ′(0) · f ′(0) · f ′(0) = 2^4 = 16
Another method is to realize that: ds dt
dt ds
− 2 st s^2 +3t^2
s^2 + 3t^2 2 st
Cool!
- If y = x^3 − 2 and x = 3z^2 + 5, then find dydz. We see that dydz = dydxdxdz , so we calculate (^) dxdy and dxdz : dy dx = 3x^2 , dx dz = 6z
so that dy dz = 3x^2 · 6 z = 3(3z^2 + 5)^2 · 6 z = 18z(3z^2 + 5)^2
- A space traveler is moving from left to right along the curve y = x^2. When she shuts off the engines, she will go off along the tangent line at that point. At what point should she shut off the engines in order to reach the point (4, 15)? The unknown in the problem is a point on the parabola y = x^2. Let’s label that point as (a, a^2 ). Now our goal is to find a. First, the line will go through both (a, a^2 ) and (4, 15), so the slope will satisfy: m = a^2 − 15 a − 4 Secondly, the line will be a tangent line, so the slope will also be m = 2a. Equating these, we can solve for a: a^2 − 15 a − 4 = 2a ⇒ a^2 − 15 = 2a(a − 4) ⇒ a^2 − 8 a + 15 = 0 ⇒ a = 3, a = 5
Since we’re moving from left to right, we would choose the smaller of these, a = 3.
- A particle moves in the plane according to the law x = t^2 +2t, y = 2t^3 − 6 t. Find the slope of the tangent line when t = 0. The slope is dydx , but we can only compute dxdt and dydt. Note however, that
dy dx
dy dt dx dt
6 t^2 − 6 2 t + 2
So, at t = 0, dydx = −3.
- Find the coordinates of the point on the curve y = (x − 2)^2 at which the tangent line is perpendicular to the line 2x − y + 2 = 0. First, recall that two slopes are perpendicular if they are negative recip- rocals (like − 3 , 13 ). The slope of the given line is 2, so we want a slope of − 12. The x that will provide this slope is found by differentiating:
y′^ = 2(x − 2) ⇒ 2(x − 2) = −
2 ⇒^ x^ =
4 from which^ y^ =^
- For what value(s) of A, B, C does the polynomial y = Ax^2 + Bx + C satisfy the differential equation: y′′^ + y′^ − 2 y = x^2
Hint: If ax^2 + bx + c = 0 for ALL x, then a = 0, b = 0, c = 0. As we did in class, compute the derivatives of y and substitute into the equation: y′^ = 2Ax + B, y′′^ = 2A so that: 2 A + 2Ax + B − 2(Ax^2 + Bx + C) = x^2 Now collect coefficients to get:
(− 1 − 2 A)x^2 + (2A − 2 B)x + (2A − 2 C) = 0 for all x
so (− 1 − 2 A) = 0, (2A − 2 B) = 0, (2A − 2 C) = 0. This gives the solution, A = − 12 , B = − 12 , C = − 12.
- If V = sin(w), w =
u, u = t^2 + 3t, compute: The rate of change of V with respect to w, the rate of change of V with respect to u, and the rate of change of V with respect to t.
dV dw
dw du = cos(w) ·
u
cos(
u) 2
u
dV dw
dw du
du dt = cos(w) ·
u · (2t + 3) = cos(
t^2 + 3t) 2
t^2 + 3t
· (2t + 3)
- Find all value(s) of k so that y = ekt^ satisfies the differential equation: y′′^ − y′^ − 2 y = 0. First, differentiate y, then substitute:
y = ekt^ ⇒ y′^ = kekt^ ⇒ y′′^ = k^2 ekt
(e) y = cot(3x^2 + 5)
y′^ = − csc^2 (3x^2 + 5)(6x) = − 6 x csc^2 (3x^2 + 5) (f) y = xcos(x) Use logarithmic differentiation: ln(y) = cos(x) · ln(x), so that 1 y y′^ = − sin(x) ln(x) + cos(x) · 1 x Multiply both sides of the equation by y, and back substitute y = xcos(x)^ to get:
y′^ = xcos(x)
− sin(x) ln(x) + cos(x) x
(g) y =
sin(
x)
y′^ =^1 2 (sin(x^1 /^2 ))−^1 /^2 cos(x^1 /^2 )^1 2 x−^1 /^2
(h)
x + √^3 y = 1 1 2 x−^1 /^2 +
y−^2 /^3 y′^ = 0
y′^ = − 3 y^2 /^3 2 x^1 /^2 (i) x tan(y) = y − 1
tan(y) + x sec^2 (y)y′^ = y′ tan(y) 1 − x sec^2 (y) =^ y
′
(j) y =
x ex 2 (x^2 + 1)^10 (Hint: Logarithmic Diff) First, we rewrite so that:
ln(y) = ln(
x ex 2 (x^2 + 1)^10 )
Use the rules of logarithms to re-write this as the sum:
ln(y) =
ln(x) + x^2 ln(e) + 10 ln(x^2 + 1) =
ln(x) + x^2 + 10 ln(x^2 + 1)
So far, we’ve only done algebra. Now it’s time to differentiate: 1 y y
′ =^1
2 ·^
x + 2x^ + 10^ ·^
x^2 + 1 ·^2 x
Simplifying, multiplying through by y:
y′^ = y
2 x
Finally, back substitute y:
y′^ =
x ex 2 (x^2 + 1)^10 ·
2 x
(k) y = sin−^1
tan−^1 (x)
This is a composition, so use the chain rule:
y′^ =
tan−^1 (x)
) 2 ·^
x^2 + 1
(l) y = ln | csc(3x) + cot(3x)| Recall that the derivative of ln |x| is (^1) x , so using the Chain Rule:
y′^ =
csc(3x) + cot(3x)
[
− csc(3x) cot(3x) · 3 − csc^2 (3x) · 3
]
Which can be simplified:
y′^ = −3 csc(3x)(cot(3x) + csc(3x)) csc(3x) + cot(3x) = −3 csc(3x)
(m) y = √− (^4) t^23 First, note that y = − 2 t−^3 /^4 so y′^ = 32 t−^7 /^4
(n) y = x 3 −^1 /x y′^ = 3−^1 /x^ + x 3 −^1 /x^ ln(3) · x−^2 (o) y = x tan−^1 (
x) Overall, use the product rule (then a chain rule):
y′^ = tan−^1 (
x)+x·
x)^2 + 1
x
== tan−^1 (
x)+
x 2(x^2 + 1)
(p) y = e^2 e
x Before putting in the values, note that this derivative will be in the form:
y′^ = e( )^ · d dx
d dx ex^ = e( )^ · 2 ( )^ ln(2) · ex
Putting in the appropriate expressions gives us:
y′^ = e^2 ex 2 e x ln(2)ex
