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Calculus I                                                  Test #2                                      April 9, 2004

Name____________________             R.  Hammack                            Score ______
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(1) Use the limit definition of the derivative to find the derivative of the function f(x) = 1/x^2.

Underscript[lim , wx] (f(w) - f(x))/(w - x) = Underscript[lim , wx] (1/w^2 - 1 ... 2 - w^2)/(x^2w^2) 1/(w - x) = Underscript[lim , wx] ((x - w) (x + w))/(x^2w^2) 1/(w - x) =

Underscript[lim , wx] -( x + w )/(x^2w^2) = -( x + x )/(x^2x^2) = (-2x)/x^4 = -2/x^3


(2) The graph of a function g(x) is shown below.  Using the same coordinate axis, sketch a graph of  y = g ' (x).


[Graphics:HTMLFiles/T2S04C_7.gif]

(3) Suppose f and g are functions for which f(3) = 10, g(3) = 8, f ' (8) = -2, and g ' (3) = 3.
Suppose also that h(x) = f(g(x)).  Find h ' (3).

h ' (x) = f ' (g(x)) g ' (x)
h ' (3) = f ' (g(3)) g ' (3) = f ' (8) 3 = (-2) (3) = 6

(4) State two things that the derivative of a function f(x) tells you.  Be specific.

    1.  f '(x) equals the slope of the tangent line to y = f (x) at the point (x,  f (x)).
    2.  f '(x) equals the rate at which the quantity  f (x) is changing with respect to x, at x.
    3.  f '(x) equals the velocity at time x of an object whose position at time x is  f (x).

(5) d/dx[ x^5 + x^3 - 3] = 5x^4 + 3x^2

(6) d/dx[ 1/x^(1/2)] =  d/dx[ x^(-1/2)] = -1/2x^(-3/2) = -1/21/x^(3/2) = -1/(2x^(1/2)^3)

(7)  d/dw[ w/(w + 1)] = (1 (w + 1) - w(1))/(w + 1)^2 = 1/(w + 1)^2

(8) d/dx[ x^2tan^2(x)    + sin(x) ] = 2x tan^2(x)    + x^22tan(x) sec^2(x) + cos(x)

(9) d/dx[   (x^5 + x^3 - 3 )^(-1)] = -(x^5 + x^3 - 3 )^(-2) (5x^4 + 3x^2) = -(5x^4 + 3x^2)/(x^5 + x^3 - 3 )^2

(10)  If     y = 1/x + 5x^2,    find (d^2y)/dx^2.

dy/dx = -1/x^2 + 10x

(d^2y)/dx^2 = 2/x^3 + 10


(11) d/dx[ sec^10(x)] =10 sec^9(x) sec(x) tan(x) = 10 sec^10(x) tan(x)


(12) d/dx[ sec(x^10) ] =sec(x^10) tan(x^10) 10x^9


(13)  d/dx[ (x + x^(1/2))^(1/2)] = d/dx[ (x + x^(1/2))^(1/2)] = 1/2 (x + x^(1/2))^(-1/2) (1 + 1/2x^(-1/2)) = (1   + &nb ... )/(2 (x + x^(1/2))^(1/2)) = (2x^(1/2)    +   1)/(4x^(1/2) (x + x^(1/2))^(1/2))

(14) d/dx[ tan(sin(cos(x))) ] = sec^2(sin(cos(x))) d/dx[ sin(cos(x)) ] = sec^2(sin(cos(x))) cos(cos(x)) (-sin(x))


(15)  Find all values of x for which the tangent line to   y = 2 sin(x)  at  (x, f(x)) has a slope of 1.

We seek those x for which dy/dx = 2cos(x) = 1,  which means  cos(x) = 1/2
The set of all such values of x is
x = π/3 + 2n π  and x = (5π)/3 + 2n π  , where n is an integer.


(16)  Find the slope of the tangent line to the graph of the equation  y/x = x y^2  at the point (1, 1).

d/dx[y/x] = d/dx[x y^2]

(y ' x   - y(1))/x^2 = (1) y^2 + x 2 y y '

y ' x   - y = x^2(y^2 + x 2 y y ')

y ' x   - y = x^2y^2 + 2x^3 y y '

y ' x   - 2x^3 y y ' = x^2y^2 + y

y ' (x   - 2x^3 y ) = x^2y^2 + y

y ' = (x^2y^2 + y)/( x   - 2x^3 y )

Now plug in the point (x,y) = (1,1) to get

y ' = (1^21^2 + 1)/( 1 - 2 (1^3 1) ) = -2 .  The tangent line has slope  -2.


(17)  Find the equation of the tangent line to the graph of y = x^3 at the point where x = 1.

Point is (1, 1^3) = (1, 1)and slope is f ' (1) = 3 (1^2) = 3.  Using the point - slope form, we get

y - 1 = 3 (x - 1)
y - 1 = 3x - 3
y = 3x - 2

(18)  A 10-foot ladder leans against a wall at an angle θ with the horizontal. The top of the ladder is y feet above the ground. If the bottom of the ladder is pushed toward the wall, find the rate y changes with respect to  θ when θ = π/3.

[Graphics:HTMLFiles/T2S04C_63.gif]

Using trig, y = 10sin(θ),  so rate of change = dy/(dθ) = 10cos (θ).
When θ = π/3, the rate of change is dy/(dθ) = 10cos (π/3) = 5feet per radian


(19)  A 10-foot ladder leans against a wall.  If the bottom of the ladder is pushed toward the wall at a rate of 2 feet per second, how quickly is the top of the ladder moving up the wall when it's 8 feet above the floor?
[Graphics:HTMLFiles/T2S04C_68.gif]

We know dx/dt = -2 (negative because x is decreasing)

We want dt/dt =?

By Pathagorean Theorem,
x^2 + y^2 = 100

d/dt[x^2 + y^2] = d/dt[100]

2xdx/dt + 2ydy/dt = 0

Since we know dx/dt = -2 and y = 8, let's plug those in now.

2x (-2) + 2 (8) dy/dt = 0

-x + 4dy/dt = 0

dy/dt = x/4

Now we just need to find x. Since y = 8 at this instant,
we can use the Pythagorean Theorem to get x = (10^2 - 8^2)^(1/2) = 36^(1/2) = 6.

ANSWER dy/dt = 6/4 = 3/2feet per second