Problems in Calculus II

Chapter 2: Review on Calculus I.

    We shall write D_x instead of (d/dx).

    Problem 1. Prove product rule of derivatives.
    Ans.: Intuitive Proof. Think of a rectangle with length u and width v. Let f = u v be its area. Suppose the length expands by du, and width by dv, then the area of the rectangle increases by a ratio du/u lengthwise and dv/v widthwise. The increase of the rectangle is due to (1) a thin strip of length v and width du, (2) a second thin strip of length u and width dv, and (3) a tiny rectangle of area du dv. Ignoring (3), df/f then = du/u + dv/v . . . . (i)
    [Common sense also tells us that the percentage increase of f is the sum of the percentage increase in u and the percentage increase of v. ]
    (i) can be simplified to
df = (u v) (du / u + dv / v) = u dv + v du . . . . (ii).
Hence df/dx = u dv/dx + v du/dx, or f ' = u v' + v u'
which is the product rule.
    Remark: You should memorise (i).

     Problem 2. Prove quotient rule of derivatives.
    Ans.: First proof. Let f = u / v, then u = f v.
From (i), then du/u = df/f + dv/v = v df / u + dv/v.
Hence df = (du/u - dv/v) (u/v) = [v du - u dv]/v²
or f ' = [v u' - u v']/v², the quotient rule.
    Ans.: Second proof. Let f = u (1/v) = area of a rectangle with length u and width 1/v. Suppose u increases by du, and v increases by dv. Then percentage increase in u is du/u, percentage increase in v is dv/v, hence percentage decrease in 1/v is dv/v.
Hence the percentage increase in f is df/f = du/u - dv/v . . . . (iii)
This simplifies to df = f [ du/u - dv/v] = (u/v) (du/u - dv/v)
= du/v - u dv/v² = [v du - u dv]/v².
    R emark: It is more meaningful to memorise (iii) as
quotient rule: f ' / f = u' / u - v' / v . . . . (iiia).

     Problem 3. Prove chain rule.
    Ans.: Proof. Let f = g(h(x)). When x increases by dx, h increases by h'(x) dx. Hence f increases by g'(h(x)) h'(x) dx because h'(x) dx is the increment of the argument of g function.
    Thus df/dx = g'(h(x) h'(x).

     Problem 4. Find D_x Ö(sin Öx).
    Ans.: Let f = Ö g where g = sin h where h = Öx. Or use your pencil and circle Öx, then circle sin Öx. Two circles means there are two intermediate functions g and h.
So D_xf = df/dg . dg/dh . dh/dx = 1/(2Ög) . cos h . 1/(2Öx)
= cosÖx / [4 Ö(sinÖx) Öx ] = cosÖx / 4 Ö[x sinÖx]
    Devious Joy. Professors rejoice in being able to sprinkle "Ö" and "²" here and there, hopefully to maximize the chance that you will stumble. You prevail only by God's grace and quench their sinister glee.

     Problem 5. Find df/dx where f = Ö[x/(1-x)]. (This is a favorite test problem.)
    Ans.: First solution. f = Ög, where g = x/(1-x).
Then D_xf = 1/2Ög . [using quotient rule] [(1-x) D_xx - x D_x(1-x)] / (1-x)²
= (½)Ö[(1-x)/x] [1-x+x]/(1-x)²] = 1/[2 Öx (1-x)^3/2]
    Second solution. f = Öx (1-x)^-1/2. Using product rule, df/dx
= Öx (-1/2)(1-x)^-3/2 (-1) + (1/2Öx) (1-x)^-1/2
[Note the "(-1)" is due to the fact that (1-x)^-1/2 = g^-1/2 where g in turn = 1-x, so that chain rule is needed, very subtle.]
= [1/ 2Öx (1-x)^3/2] [x +(1-x)] = 1/ 2Öx (1-x)^3/2.

     Problem 6. State and prove the Fundamental Theorem of Calculus.
    Ans.: Theorem: Let f(t) be a continuous function for a £ t £ b. Then D_x ò_c^x f(t) dt = f(x), for any c, x such that a £ c £ x £ b.
    Proof. Let F(x) = ò_c^xf(t) dt. It is the area between the curve y = f(t) and the t-axis from t=c to t=c. Hence F(x+dx)-F(x) is the area dA under the curve from t = x to t = x+dx. When dx is very small compared to x, the area dA is that of a thin strip, whose difference from the area of the rectangle with height x and width dx is something whose ratio with dA approaches zero as dx approaches zero. Hence dA = (f(x) + e)dx, and lim dA/dx = f(x) because e also ®0, due to continuity of f at the point t = x. Hence D_xf(x) = lim dA/dx = f(x).
    Warning: Do not confuse x with t. You must quote the statement of the theorem precisely as shown. Don't even change a preposition or an adverb.

     Problem 7. Find D_x ò_{sin x}^{tan x} Öt dt.
    Ans.: It = D_x [ ò_c^{tan x} Öt dt ] = D_x [ ò_c^{tan x} Öt dt + ò_{sin x}^c Öt dt ]
= D_x [ ò_c^{tan x} Öt dt - ò_{sin x}^c Öt dt ]
=, using chain rule, Ö(tan x) - Ö(sin x).

    Problem 8. Prove that if 0 < q < 90° then sinq < q < tanq, and lim_q®0 (sinq /q) = 1.
    Ans.: Let AB be an arc of a unit circle with center O such that ÐAOB = q. Let D be the projection of A upon OB. Let the tangent line at B cut OA produced at C.
Then area of DAOB < area of sector AOB < area of DOBC.
Hence (½)OB.AD < (½)OB²q <(½)OB.BC
OB = 1. \ sinq < q < tanq.
    Divide by sinq, we get 1 < q/sinq < 1/cosq.
Hence 1 £ lim_q®0 (q/sinq) £ 1/lim_q®0 cosq,
Now lim cosq = 1, hence lim(q/sinq) is "squeezed" to 1 also.
Hence lim (sinq /q) = 1 / lim (q/sinq) = 1 / 1 = 1. (QED)
    Warning. Some would start the proof from asserting that AD < arc AB < BC. It is only common sense that chord AB < arc AB, but a good proof using Euclidean geometry is not obvious. The other fact that arc AB < length BC is even more difficult to prove. In fact, since we have sinq < q < tanq, we can deduce that AD < arc AB < line BC. This fact is a consequence, not a cause of the theorem.

    Problem 9. Prove D_x sin x = cos x, and D_x tan x = sec² x.
    Ans.: D_xsin x = lim_h®0 [sin(x+h)-sin x]/h
= lim_h®0 2 cos (x+h/2) sin h /h = [lim 2 cos(x+h/2)] [lim (sin h /h)]
cos x . 1 = cos x.
     Again, lim_h®0 [tan(x+h)-tan x]/h
= lim [2 sin(x+h)cos x - 2 cos(x+h)sin x]/ 2 h cos(x+h) cos x
= lim [sin (2x+h) + sin h - sin(2x+h) + sin h]/ 2 h cos(x+h) cos x
= 2 sin h / 2 h cos(x+h) cos x = 1/(cos x cos x) = sec²x.

    Problem 10. Prove that D_xÖx = 1/2Öx.
    Ans.: (Tricky!) D_xÖ = lim_h®0(1/h) [Ö(x+h) -Öx]
= lim [Ö(x+h)-Öx] [Ö(x+h) +Öx] / h[Ö(x+h)+Öx]
= lim [ x+h - x] / h[Ö(x+h)+Öx] = lim 1/[Ö(x+h)+Öx]
= 1/ (2Öx).

    Problem 11. Prove Binomial Theorem.
    Ans.: Binomial Theorem says
(1+x)ⁿ = 1 + ⁿC₁x + ⁿC₂x² + . . . + ⁿC_rx^r + . . . + xⁿ . . . . (11.1),
where n is an integer > 0, and each ⁿC_r = n (n-1) (n-2) . . . (n-r+1) / r ! are called Binomial coefficient, or "number of combinations of r things out of n".
    It is important to know that ⁿC₀ = 1, ⁿC₁ = n, and ⁿC₂ = n(n-1)/2.
    An important property is the Pascal Triangle property:
ⁿC_r-1 + ⁿC_r = ⁿ⁺¹C_r. Proof:
LHS = n (n-1) . . . (n-(r-1)+1) / (r-1) ! + n (n-1) . . . (n-r+1) / r !
= n (n-1) . . . (n-r+2) [ r + (n-r+1) ] / r !
= (n+1) n . . . (n-r+2) / r ! = ⁿ⁺¹C_r.
Now we can prove Binomial theorem by induction:
Case n = 1: LHS = 1 + x. RHS = 1 + ¹C₁x = 1 + x. Hence LHS = RHS.
    Assume truth for case n, i.e., (11.1) is true.
Then for case n+1, (1+x)ⁿ⁺¹ = (1+x) (1+x)ⁿ
= (1+x)ⁿ = 1 + ⁿC₁x + ⁿC₂x² + . . . + ⁿC_rx^r + . . . + xⁿ
+ x + ⁿC₁x² + ⁿC₂x³ + . . . + ⁿC_rx^r+1 + . . . + xⁿ⁺¹
= 1 + x(ⁿC₁ + ⁿC₀) + x²(ⁿC₂+ⁿC₀) + . . . + xⁿ⁺¹
= 1 + ⁿ⁺¹C₁x + ⁿ⁺¹C₂x² + . . . + xⁿ⁺¹.
Conclusion: Since the theorem is true for case 1, it is true for case 2, since it is true for case 2, it is true for case 3, etc. Hence by Mathematical Induction, it is true for all integers n > 0.
    Remark: If n is any positive real number, and |x| < 1, then the RHS of the theorem becomes an infinite power series, and it will converge only if |x| < 1. The proof depends on Taylor's Theorem. But assuming its truth, we see that
Ö(1+x) = 1 + (½)x -x²/8 + x³/16 -(5/128)x⁴ + - . . .
called the Square Root power series.

     Problem 12. Prove that D_x xⁿ = n x^n-1.
    Ans.: Case 1: n is integer > 0. Then
D_x xⁿ = lim_h®0 (1/h)[(x+h)ⁿ-xⁿ ]
= lim (1/h) [n x^n-1h + (½)n(n-1)x^n-2h² + . . .]
= lim [n x^n-1 + (½)n(n-1)x^n-2h + . . .]
= n x^n-1 + 0 + 0 + . . . (finite terms)
= n x^n-1.
    Case 2: n = p/q, where p,q > 0 are integers. Let f = x^p/q.
Then f^q = x^p. Hence D_x(f^q) = D_x(x^p).
\ using chain rule for LHS, q f^q-1 df/dx = p x^p-1
\ q x ^p(q-1)/q df/dx = p x^p-1.
\ df/dx = (p/q) x^p-1-p(q-1)/q = (p/q) x^(p/q)-1
\ df/dx = n x^n-1, when n is fractional.
    Case 3: n < 0. Let n = -m. Let f = xⁿ = x^-m. Then 1 = x^m f.
Using product rule, we get 0 = m x^m-1 f + x^m df/dx.
Hence df/dx = - m x^m-1-m f = -m x^-1 x^-m = n x^n-1.
    Case 5: n is irrational. This requires theory of limiting functions from advanced calculus, so we now take it by faith.