MATH 0220 EXAM II NAME1. Differentiate each of the following. It’s not necessary to simplify your answers.(a) f(x) =3√1 + x + 3x2f0(x) = −3(1 + 6x)2(1 + x + 3x2)3/2(b) s(t) = arctan (2t)s0(t) =21 + 4t2(c) y =tan x1 + cos xy0(x) =(1 + cos x) sec2x + tan x sin x(1 + cos x)2(d) g(x) = ln (x + e−3x)g0(x) =1 − 3e−3xx + e−3x(e) f(x) =5(x2− 4)3f0(x) = −30x(x2− 4)4(f) y = xcos xy0(x) = xcos x−sin x ln x +cos xx2. (a) Determine the equation of the tangent line to the curve:x2sin y + 3x + 2y = 6 at the point (2, 0).2x sin x + x2cos ydydx+ 3 + 2dydx= 00 + 4dydx+ 3 + 2dydx= 0dydx= −12Therefore, the equation of t he tangent line is y = −12(x − 2)(b) Use your answer in part (a) to approximate y when x = 2.4y ≈ −0.5(2.4 − 2) = −0.5(0.4) = −0.23. Find the point(s) on the parabola y = x2that is (are) closest to the point (0, 1).Minimize Distance to (0,1) under constraint point on parabola.D =q(x − 0)2+ (y − 1)2y = x2D(y) =qy + (y − 1)2D0(y) =1 + 2(y − 1)2qy + (y − 1)22y − 1 = 0y =12x = ±s12Points are:s12,12and−s12,12.This could have been done replacing y with x2. Then:D(x) =qx2+ (x2− 1)2D0(x) =2x + 2(x2− 1)(2x)2qx2+ (x2− 1)22x + 2(x2− 1)(2x) = 04x3− 2x = 02x(2x2− 1) = 0x = 0, x = ±s12x = 0 is a local max. Therefore, the points a r es12,12and−s12,12.4. A lighthouse is 100 meters from a straight shoreline. The light turns at a rate of10 revolutions per minute (20π radians/minute), and shines a moving spot of lightalong the shore. How fast is the spo t of light moving when it’s 100 meters f r om thepoint on t he shore which is nearest the lighthouse? Be sure to include units in youranswer.Draw a straight shoreline. Now draw the different points of light on this straightshoreline and label one x. Put the light 100 meters from the shoreline. D r aw thelight from the lighthouse to some of the points, namely the one labeled x and theone that is exactly 100 meters from the lighthouse. What you have is a r ig ht trianglewith the adjacent side 100 meters and the opposite side x meters. θ is the angle ofrotation of the light. Therefore,tan θ =x100sec2θdθdt=1100dxdt2(20π)(100)m/min =dxdt5. Use Newton’s Method once to estimate the solution to the equation x = e−x. (Notethat the plot of f(x) = x − e−xis shown below.)0 1xLet x0= 0. Then f(x0) = f(0) = −1 and since f0(x) = 1+e−x, f0(x0) = f0(0) = −1and we have:x1= 0 −(−1)2=126. Determine each of the following limits. Show your work.(a) limx→0sin xex− 1L0Rule= limx→0cos xex=11= 1.(b) limx→0ex− 1cos x=01= 0.(c) limx→0(1 + 2x)1/xLet y = (1 + 2x)1/x. Then ln y =ln((1 + 2x)xandlimx→0ln y = limx→0ln (1 + 2x)xL0Rule= limx→021+2x1= 2.Now if ln y → 2, then y →