Exam 2 Review Solutions1. True or False, and explain:(a) The derivative of a polynomial is a polynomial.True. A polynomial is a function of the form a0+a1x+a2x2+. . . anxn,and its derivative will also have integer powers of x by the Power Rule,nxn−1.(b) If f is differentiable, thenddxpf(x) =f0(x)2√f(x)True:ddxpf(x) =ddx(f(x))1/2=12(f(x))−1/2f0(x) =f0(x)2pf(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, whichis not the reciprocal of the secant.For extra practice, to get the formula for the derivative of y =sec−1(x):sec(y) = xFrom this, draw a right triangle with one acute angle labelled y, thehypotenuse x and the adjacent length x. This gives the length of theside opposite:√x2− 1. Now differentiate:sec(y) tan(y)dydx= 1From the triangle, sec(y) = x and tan(y) =√x2− 1, so:dydx=1x√x2− 1(d)ddx(10x) = x10x−1False. The Power Rule can only be used for xn, not ax. The derivativeis 10xln(10).(e) If y = ln |x|, then y0=1x.TRUE. To see this, re-write the function:y =ln(x) if x > 0ln(−x) if x < 0⇒ y0=1xif x > 01−x· (−1) if x < 0From which we see that y0=1x, x 6= 0.(f) The equation of the tangent line to y = x2at (1, 1) is:y − 1 = 2x(x − 1)1False. This is the equation of a parabola, not a line. The derivative,y0= 2x gives a formula for the slope of the tangent line, and isnot the slope itself. To get the slope, we evaluate the derivative atx = 1, which gives y0= 2. The slope of the tangent line is thereforey − 1 = 2(x − 1).(g) If y = e2, then y0= 2eFalse. e2is a constant, so the derivative is zero.(h) If y = |x2+ x|, then y0= |2x + 1|.False. First, rewrite y, then differentiate:y =x2+ x if x ≤ 0 or x ≥ 1−(x2+ x) if 0 < x < 1⇒ y0=2x + 1 if x < 0 or x > 1−(2x + 1) if 0 < x < 1Compare this to |2x + 1|:|2x + 1| =2x + 1 if x ≥ −1/2−(2x + 1) if x < −1/2By the way, we also note that y is NOT differentiable at x = 0 orat x = 1 by checking to see what the derivatives are approaching asx → 0 and as x → 1.(i) I f y = ax + b, thendyda= xTrue.dydameans that we treat a as an indep e ndent variable, and x, bas constants.2. Find the equation of the tangent line to x3+ y3= 3xy at the point (32,32).We need to find the slope,dydxx=3/2,y=3/23x2+ 3y2y0= 3y + 3xy0⇒ y0(3y2− 3x) = 3y − 3x2⇒ y0=y − x2y2− xSubstituting x = 3/2, y = 3/2 gives y0= −1, so the equation of thetangent line is y − 3/2 = −1(x − 3/2)3. If f(0) = 0, and f0(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:f0(f(f(f(x)))) · f0(f(f(x))) · f0(f(x)) · f0(x)Now subsitute x = 0 and evaluate:f0(f(f(f(0)))) · f0(f(f(0))) · f0(f(0)) · f0(0)f0(0) · f0(0) · f0(0) · f0(0) = 24= 1624. If f(x) = 2x + ex, find the equation of the tangent line to the INVERSEof f at (1, 0).First, we verify that (0, 1) is on the graph of f :f(0) = 2 · 0 + e0= 1We know that, if f0(0) = m, thendf−1dxx=1=1m.Now, f0(x) = 2 + ex, so f0(0) = 3. Therefore, the slope of the tangent lineto the inverse of f at x = 1 is13, and the equation is then:y − 0 =13(x − 1) or y =13x −135. Derive the formula for the derivative of y = cos−1(x) using implicit differ-entiation.First, cos(y) = x, so that implicit differentiation gives −sin(y)y0= 1, soy0= −1/ sin(y). Now to convert this back to x, draw a right triangle withy as one of the acute angles. Label the adjacent side as x, hypotenuse as1, so the length of the side opposite is√1 − x2. This gives:y0=−1√1 − x26. Find the equation of the tangent line to√y + xy2= 5 at the point (4, 1).Implicit differentiation gives:12y−1/2y0+ y2+ 2xyy0= 0Now we could solve for y0now, or substitute x = 4, y = 1:121−1/2y0+ 12+ 2(4)(1)y0= 0 ⇒ y0= −2/17The equation of the tangent line is y − 1 =−217(x − 4)7. If s2t + t3= 1, finddtdsanddsdt.The notationdtdsmeans that we are treating t as a function of s. Therefore,we have:2st + s2dtds+ 3t2dtds= 0 ⇒dtds=−2sts2+ 3t2For the second part, we have two choices. One choice is to treat s as afunction of t and differe ntiate:2sdsdtt + s2+ 3t2= 0 ⇒dsdt=−(s2+ 3t2)2st3Another method is to realize that:dsdt=1dtds=1−2sts2+3t2= −s2+ 3t22stCool!8. If y = x3− 2 and x = 3z2+ 5, then finddydz.We see thatdydz=dydxdxdz, so we calculatedydxanddxdz:dydx= 3x2,dxdz= 6zso thatdydz= 3x2· 6z = 3(3z2+ 5)2· 6z = 18z(3z2+ 5)29. A space traveler is moving from le ft to right along the curve y = x2.When she shuts off the engines, she will go off along the tangent line atthat point. At what point should she shut off the engines in order to reachthe point (4, 15)?The unknown in the problem is a point on the parabola y = x2. Let’slabel that point as (a, a2). Now our goal is to find a.First, the line will go through both (a, a2) and (4, 15), so the slope willsatisfy:m =a2− 15a − 4Secondly, the line will be a tangent line, so the slope will also be m = 2a.Equating thes e, we c an solve for a:a2− 15a − 4= 2a ⇒ a2− 15 = 2a(a − 4) ⇒ a2− 8a + 15 = 0 ⇒ a = 3, a = 5Since we’re moving from left to right, we would choose the smaller of these,a = 3.10. A particle moves in the plane according to the law x = t2+2t, y = 2t3−6t.Find the slope of the tangent line when t = 0.The slope isdydx, but we c an only computedxdtanddydt. Note however, thatdydx=dydtdxdt=6t2− 62t + 2So, at t = 0,dydx= −3.411. Find the coordinates of the point on the curve y = (x − 2)2at which thetangent 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:y0= 2(x − 2) ⇒ 2(x − 2) = −12⇒ x =74from which y =11612. For what value(s) of A, B, C does the polynomial y = Ax2+ Bx + Csatisfy the differential equation:y00+ y0− 2y = x2Hint: If ax2+ 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 theequation:y0= 2Ax + B, y00= 2Aso that:2A + 2Ax + B − 2(Ax2+ Bx + C) = x2Now collect coefficients to get:(−1 − 2A)x2+ (2A − 2B)x + (2A − 2C) = 0 for all xso (−1 −2A) = 0, (2A −2B) = 0, (2A −2C) = 0. This gives the solution,A = −12, …