983176648 test 3 revised

Test 3
Full name
University Affiliation

a) A graph of the function is given below

Fill in the following table with approximate values (if they exist)
2 4 6 8 10 12 14
1 0 1 (b) For the function
find and an equation of the tangent line to the graph of the above function at
In the questions c to h , find
(C )
y = x3 – x2 – x5/3
= 3×2 – 2x – 5/3x
(d) y=x3 ln(1+x2)
You may use
(e)
y’= 2+cosx+ ddxsinx-sinx. ddx(2+cosx)(2+cosx)2y’= 2+cosx.cosx-sinx. (-sin⁡(x))(2+cosx)2y’= 2+cosx+cos2 x+sin2(x)(2+cosx)2cos2 x+sin2x=1sin2x=1- cos2 xy’= 2+cosx+cos2 x+1- cos2 x(2+cosx)2y’= 1+ 2cosx(2+cosx)2(f)
y=sin-1(ex)siny=(ex)cosyy’=(ex)y’=excosyy=ex1-e2x(g) Find if
i.e.
= ddxsin3(2x)=3sin22x.cos2x.2(2) (a) Find an equation of the tangent line to the graph of
y=2xcosx at 0,1
Use Geogebra or any other tool to draw a graph of y=2xcosx and that of the tangent line to the graph of y=2xcosx at 0,1 in the same window.

(b) Find an equation to the tangent line to the graph of
at
Use geogebra or any other tool to draw a graph of and that of the tangent line to the graph of at in the same window.

(c) We are given the graphs of and below. Find , i.e. the derivative of at 1.

(g∘f)′(1)=g′(f(1))f′(1)
=g′(e)f′(1)
=(2e−4)(e)
=2e2−4e
(d) Find the differential of . Find , an equation of the tangent line to the graph of at

y = mx + c
y = 2 x = 2 m = 0.42
c = 1.16
y = 0.42x+c
Two cars start moving from the same point. One travels east at a constant rate of 60 miles per hour and the other travels north at a constant rate of 70 miles per hour.

Find the rate at which the distance between the cars is changing 1 hour later. You may round your answer to two digits after the decimal sign.
Distance = Rate x Time
A = 60 x 1 = 60 Miles
B = 70 x 1 = 70 Miles
3619508953500
70 z
60
x2+y2=z23600 + 4900 = z2
Z = 92.20
X = 60
Y = 70
Z = 92.20
dx/dt = 60dy/dt = 70dz/dt
x2+y2=z22x. dx/dt + 2y. dy/dt = 2z. dz/dt
x. dx/dt + y. dy/dt = z. dz/dt
60(60) + 70(70) = 92.2. dz/dt
8500/92.2 = dz/dt
= 92.20
(e) Find if and find an equation of the tangent line to the graph of at
dydxx+y=1+y’cosxyy+xy’=1+y’ ycosxy-1=y’-xcos(xy)y’
1-xcos(xy)y’
y’=ycosxy-11-xcos(xy)

x=0y=0y=mx+cc=0(f) Find an equation of the tangent line to the graph of

at
dydx= x(25-4×2-4y2)y(25+4×2+4y2)3(25-4.32-4.12)1(25+4.32+4.12)==-913y=-913x+4013
(g) Use Newton’s method to find the solution of
by using the following steps
(i) Find, where
xn+1=xn- f(x)f'(x)f’x= 6x²+5×0 f(x)f'(x)xn+1=xn- f(x)f'(x)0-651.2x=1.2ii) Find, where
xn+1=xn- f(x)f'(x)f’x= 6x²+5×0 f(x)f'(x)x0+1=x0- f(x)f'(x)11110.0909x=0.0909iii)x0 = 3
iv)Find, where
xn+1=xn- f(x)f'(x)f’x= 6x²+5×0 f(x)f'(x)x0+1=x0- f(x)f'(x)363591.06781.067811.8411810411.841181043