Solutions to Practice Final Questions (Math 31B)1N. B. Sometimes you will see that I write “log” when I meant the natural logarithm. I willtry my best to resist this tendency, but in case you do see “log” without the base writtennext to it, that means the natural logarithm, “ln”.1. Inverse Function Theorem Problems(a) Use the inverse function theorem to compute f0(x) if f(x) = arctan(x). Notethat the domain of tan(x) is given to be (−π/2, π/2). What is the domain ofarctan(x)?Solution: Define g(x) := tan(x). Then the derivative of the inverse isf0(x) =1g0(f(x)).Plug in to getf0(x) =1sec2(arctan(x)).But then, if θ = tan(x), then sec2(arctan(x)) = sec2(θ) = 1 + tan2(arctan(x)) =1+x2, so f0(x) = 1/(x2+1). As for the domain of arctan, recall that the domain ofarctan(x) is equal to the range of tan(x) (There is a special relationship betweenthe domain/range of a function and the domain/range of its inverse. What is it?).The range of tan(x) is all real numbers (or, equivalently (−∞, ∞).Answer: f0(x) = 1/(x2+ 1), (−∞, ∞)(b) Use the inverse function theorem to compute f0(x) if f(x) = arcsec(x). Note thatthe domain of sec(x) is given to be [0, π/2) ∪ (π/2, π]. What is the domain ofarcsec(x)?Solution: Similar drill, except that we let g(x) := sec(x). Recall that g0(x) =sec(x) tan(x). Applying the theorem on inverse gives youf0(x) = 1/(x√x2− 1).As for the domain of arcsec, recall that the range of sec is the reciprocal of [−1, 1].So it is (−∞, −1] ∪ [1, ∞).Answer: f0(x) = 1/(x√x2− 1), (−∞, −1] ∪ [1, ∞).(c) Suppose that f is an invertible function on an open interval containing 5. Usethe inverse function theorem to compute the equation of the line tangent to thegraph of the function f−1(x) at the point (3, 5) if f(5) = 3 and if f0(5) = 4.Solution: Use the inverse theorem to get(f−1)0(3) =1f0(f−1(3))=1f0(5)=14.1Last updated: 2012-12-05 23:01:221So the equation of the tangent line is y =14(x − 3) + 5.Answer: y =14(x − 3) + 5.2. Differentiating logs and exponential functions(a) Compute the derivative of the functionf(x) = log5(cos2(x) + x2+ 1).Solution: Use the logarithmic rule to re-write f in terms of the natural logarithm.Then apply the chain rule to getf0(x) =ddxlog(cos2(x) + x2+ 1)log 5=1log 5·ddx(cos2(x) + x2+ 1)cos2(x) + x2+ 1=−2 cos(x) sin(x) + 2xlog 5(cos2(x) + x2+ 1).(b) Compute the derivative of the functionf(x) = xx2+sin(x).Solution: Take the logarithm on both sides, differentiate both sides, and get thederivative (a.k.a. logarithmic differentiation!).(c) Compute the derivative of the functionf(x) = xxx.Solution: Apply natural logarithm on both sides twice, and differentiate bothsides. This is a “double logarithmic differentiation” problem.(d) Suppose that x > 1. Compute the derivative of the functionf(x) = logxx2.Solution: Since f(x) = 2, it follows that f0(x) = 0.(e) Suppose that x > 1. Compute the derivative of the functionf(x) = logx(sin(x) + 2).Solution: Re-write f(x) using ln, and then use the quotient rule.28. Improper integrals(a) Is the following integral convergent or divergent:Z∞0x21 + x4dx.Solution: It looks likex2x4=1x2is going to be a good choice for the comparisontest. But note thatR1x2dx is divergent if integrated over (0, a]. That causes someproblems. To get around, we shall split the integral into two pieces:Z10x21 + x4dx +Z∞1x21 + x4dx.(Remark: Any choice of a will work.)The first integrand will have a definite value – note that the function is definedeverywhere in [0, 1], and the given function is continuous, so this portion won’taffect the convergence or divergence of the original integral in any way. Do thecomparison test for the second integrand:Z∞1x21 + x4dx <Z∞11x2dx = limR→∞−R−1+ 1 = 1,so the given integral converges.(b) ComputeZ∞21x(ln(x))2dx.Solution: Apply substitution. Since this integral involves substitution, I willobtain the indefinite integral first, and then plug in endpoints. Let u := ln x.Then du = dx/x.Z1x(ln(x))2dx =Zduu2= −1u+ C = −1log x+ C.Now time to compute the definite integral.Z∞21x(ln(x))2dx = limR→∞−1log R+1log 2=1log 2.Answer: 1/ log 2.3(c) Is the following integral convergent or divergent:Z10sin(x)(x2+ 1)x√xdx.Solution: The key is to notice sin(x) < x in [0, 1]. First, in [0, π/2], the sinefunction is an increasing function, and sin(0) = 0 < 1. But we know that sin xwon’t hit 1 until x = π/2, and π/2 is greater than 1. Alternately, one case usethe Taylor series of the sine function. Recall thatsin(x) = x −x33!+x55!−x77!+ ··· +(−1)nx2n+1(2n + 1)!+ ··· .If you apply the ratio test, you see that the magnitude of each term keeps de-creasing. Thus, sin(x)−x will remain negative if 0 ≤ x ≤ 1, as −x33!+x55!+··· willremain negative. This shows that sin(x) < x. (Credit to Stephanie Lewkiewiczfor pointing this out to me). Now there are two ways to proceed, both of whichare similar to each other but nonetheless slightly different.Method 1 (Again, credit to my fellow TA, Stephanie Lewkiewicz)We start by splitting the integral, and bound the function by above. Note thatthe following holds for all x ∈ (0, 1]:(x2+ 1) sin(x)x√x=√x sin(x) +sin(x)x·1√x<√x sin(x) +1x1/2.So we haveZ10(x2+ 1) sin(x)x3/2<Z10√x sin(x) dx +Z10dxx1/2.The first integrand is irrelevant – the product of two continuous functions iscontinuous, so the first integrand has a definite value. It all boils down to thesecond integrand. But we know that integral is convergent, by the p-integral test(Or, if you don’t want to memorize the test, just evaluate. This shouldn’t takelong.). Therefore, the given integral is convergent.Method 2Again, since sin(x) < x for x ∈ [0, 1], it follows that sin(x)/x < 1 for x ∈ (0, 1].Therefore, we haveZ10(x2+ 1) sin(x)x3/2dx =Z10sin(x)x·x2+ 1x1/2dx <Z10x2+ 1x1/2dx <Z102x1/2dx.The last inequality holds, since if 0 ≤ x ≤ 1 then 1 ≤ x2+ 1 ≤ 2. Apply thep-integral test to the very last integrand (or just evaluate) to conclude that theoriginal integral is convergent.4(d) Is the following integral convergent of divergent:Z∞21x(ln(x))dx.Solution: Rather than performing the comparison test, we can directly integratethis: it’s pretty clear that we need to do the substitution. Let u := ln x. Thendu = dx/x. So we haveZ1x ln(x)dx =Zduu= ln u + C = ln(ln x) + C.Now we evaluate the definite integral:Z∞21x(ln(x))2dx = limR→∞ln(ln R) − ln(ln 2) = ∞.Therefore, the given