MATH 18.01 - MIDTERM 1 - SOME REVIEW PROBLEMS WITH SOLUTIONS 18.01 Calculus, Fall 2014 Professor: Jared Speck Problem 1. Compute the second derivative of the function f (x) = arctan x. x Problem 2. Compute the derivative of the function f (x) = sin(x)xx . Problem 3. Compute limx→π/2 L’Hôpital’s rule. cos(3x) . cos(x) Problem 4. Compute the derivative dy dx At this point in the course, you are forbidden from using for the curve x2 + y 3 + xy = 7 at the point (x, y) = (2, 1). Problem 5. Find the equation for the tangent line to y = ln(x/3) at x = 3e. Problem 6. Consider the function sin x − 2x f (x) = 0 cos x−1 x2 if x < 0, if x = 0, if x > 0. Is f (x) continuous at x = 0? If not, then is the discontinuity removable? Problem 7. Prove the quotient rule (u/v)0 = (u0 v −uv 0 )/v 2 using only the definition of a derivative. Problem 8. Let r be a real number. Compute limh→0 derivative. 1 (1+2h)r −1 h by interpreting this limit as a 2 MATH 18.01 - MIDTERM 1 - SOME REVIEW PROBLEMS WITH SOLUTIONS Solutions Problem 1. Compute the second derivative of the function f (x) = arctan x. Solution: d 1 arctan x = , (you should know how to prove this), dx 1 + x2 d2 1 d −2x arctan x = = . dx2 dx 1 + x2 (1 + x2 )2 x Problem 2. Compute the derivative of the function f (x) = sin(x)xx . Solution: We first compute the derivative of xx using logarithmic differentiation: y = xx , ln y = ln(xx ) = x ln x, y0 = 1 + ln x, y y 0 = (1 + ln x)xx . x We now compute the derivative of xx : x z = xx , x ln z = ln(xx ) = xx ln x, d d 1 z0 = ln x xx + xx ln x = ln x(1 + ln x)xx + xx , z dx dx x 1 x . z 0 = xx ln x(1 + ln x)xx + xx x x Finally, we compute the derivative of sin(x)xx using the product rule: d d x x x d sin(x)xx = xx sin(x) + sin(x) xx dx dx dx xx = x cos x + sin(x)x Problem 3. Compute limx→π/2 L’Hôpital’s rule. Solution: cos(3x) . cos(x) xx x ln x(1 + ln x)x + x x1 x . At this point in the course, you are forbidden from using MATH 18.01 - MIDTERM 1 - SOME REVIEW PROBLEMS WITH SOLUTIONS 3 cos(3x) − cos(3π/2) cos(3x) = lim x→π/2 cos(x) − cos(π/2) x→π/2 cos(x) cos(3x) − cos(3π/2) x − π/2 = 3 lim x→π/2 3x − 3π/2 cos(x) − cos(π/2) cos(h) − cos(3π/2) 1 = 3 lim × cos(h)−cos(π/2) h→3π/2 h − 3π/2 limh→π/2 lim h−π/2 1 , cos(u)|u=π/2 1 = 3(− sin(3π/2)) × − sin(π/2) = 3(1)(−1) = −3. =3 d cos(u)|u=3π/2 × du Problem 4. Compute the derivative dy dx d du for the curve x2 + y 3 + xy = 7 at the point (x, y) = (2, 1). Solution: 2x + 3y 2 y 0 + xy 0 + y = 0, y0 = − 2x + y 5 = − = −1. 2 3y + x 5 Problem 5. Find the equation for the tangent line to y = f (x) = ln(x/3) at x = 3e. Solution: f (3e) = ln e = 1, 1 1 1 f 0 (x) = × = , x/3 3 x 1 f 0 (3e) = . 3e Tangent line (with x0 = 3e): y − f (x0 ) = f 0 (x0 )(x − x0 ), 1 y − 1 = (x − 3e). 3e Problem 6. Consider the function sin x − 2x f (x) = 0 cos x−1 x2 if x < 0, if x = 0, if x > 0. 4 MATH 18.01 - MIDTERM 1 - SOME REVIEW PROBLEMS WITH SOLUTIONS Is f (x) continuous at x = 0? If not, then is the discontinuity removable? Solution: We recall the following important limits, which have been previously investigated in this course: sin x 1 =− , x→0 2x 2 cos x − 1 1 lim+ =− . 2 x→0 x 2 Therefore, since limx→0 f (x) = − 12 6= f (0), f (x) is not continuous at x = 0. However, if we redefine f by setting f (0) = − 12 , then for the new f, we have limx→0 f (x) = f (0). Thus, the redefined f is continuous (and hence the original f has a removable discontinuity at x = 0). lim− − Problem 7. Prove the quotient rule (u/v)0 = (u0 v −uv 0 )/v 2 using only the definition of a derivative. Solution: Given a point x and a small number ∆x, we define ∆u = u(x + ∆x) − u(x), ∆v = v(x + ∆x) − v(x). Since u, v are differentiable (and therefore continuous), we have lim ∆u = 0, ∆x→0 lim ∆v = 0, ∆x→0 ∆u = u0 , ∆x→0 ∆x ∆v = v0. lim ∆x→0 ∆x lim Using the above facts, we have that u 0 v = lim ∆x→0 = lim ∆x→0 = lim u+∆u v+∆v − ∆x u v (u+∆u)v−(v+∆v)u v(v+∆v) ∆x v∆u−u∆v v(v+∆v) ∆x ∆u v ∆v u = lim − lim ∆x→0 ∆x v(v + ∆v) ∆x→0 ∆x v(v + ∆v) 0 0 uv vu = 2 − 2 v v u0 v − uv 0 = . v2 ∆x→0 MATH 18.01 - MIDTERM 1 - SOME REVIEW PROBLEMS WITH SOLUTIONS Problem 8. Let r be a real number. Compute limh→0 derivative. (1+2h)r −1 h by interpreting this limit as a Solution: (1 + 2h)r − 1 d = (1 + 2x)r |x=0 = r(1 + 2x)r−1 × 2|x=0 = 2r. h→0 h dx lim 5