Math 1310-004
Second Midterm Sample
1.
(a) Use logarithmic differentiation to find the derivative of: f ( x ) = a
√
2 + x 2 a 2 − x 2
(Do not try to simplify your answer!)
(b) Compare the derivatives of f ( x ) = ln(tan
− 1
( x )) and g ( x ) = tan
− 1
(ln( x ))
2.
(a) Find a linear approximation of the function: y =
√
3 x near x = 27
√
3 and use it to find an approximate value for 28.
(b) Find a linear approximation of the function: f ( x ) = x
100 near x = 1 and use it to find an approximate value for (1 .
001)
100
.
3.
A woman 5 ft tall is walking away from a street light mounted at the top of a 10 foot pole. If the woman is walking at 5 ft per second, how fast is the length of her shadow increasing?
4.
(a) Find the absolute maximum and minimum values of f ( x ) = x
3
− x on the interval [0 , 3]
(b) Find the absolute maximum and minimum values of: f ( x ) = x − ln( x ) on the interval (0 , ∞ )
In both cases, use the second derivative test to explain your answer.
5.
Find the following limits: lim x → 0
+
( e x x
− 1)
2
, lim x → 0 sin( x ) − x x 3
, lim x → 0
+ x
2 ln( x ) , lim x →∞
(1 + n x
) mx
6.
A cylinder without a top is to contain a fixed volume V of liquid.
What is the minimum amount of material necessary to construct the cylinder? What is the ratio of the radius to the height of the cylinder that minimizes the amount of material?
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Math 1310-004
The Cheat Sheet
Rules for Differentiating Combinations of Functions.
( cf )
0
= cf
0
, ( f + g )
0
= f
0
+ g
0
, ( f − g )
0
= f
0
− g
0
( f g )
0
= f
0 g + f g
0
, (1 /g )
0 g
0
= − g 2
, ( f /g )
0
= f
0 g − f g
0 g 2
Let u = g ( x ) and F ( x ) = f ( g ( x )) = f ( u ). Then F
0
( x ) = f
0
( u ) · g
0
( x ).
The Basic Derivatives (so far).
d dx
( c ) = 0 , d dx
( x n
) = nx n − 1 as long as n = 0 d dx
( e x
) = e x
, d dx ln ( x ) = d dx
( c x
) = c x
1 x
, d dx log a ln( c ) for all
( x ) = x
1 ln( a c >
)
0 d d dx dx d
(sin( x )) = cos( x ) ,
(tan( x )) = sec
2 d dx
(sin
− 1
( x )) =
( x ) , dx
(sec( x )) = sec( x ) tan( x ) ,
1
√
1 − x 2
, d dx
(cos( x )) = − sin( x ) d
(cot( x )) = − csc
2
( x ) dx d dx
(csc( x )) = − csc( x ) cot( x ) d dx
(cos
− 1
( x )) = −
1
√
1 − x 2 d dx
(tan
− 1
( x )) =
1 +
1 x 2
Some Trig Identities.
sin
2
( x ) + cos 2 ( x ) = 1 sin( x + y ) = sin( x ) cos( y ) + cos( x ) sin( y ) sin( x − y ) = sin( x ) cos( y ) − cos( x ) sin( y ) cos( x + y ) = cos( x ) cos( y ) − sin( x ) sin( y ) cos( x − y ) = cos( x ) cos( y ) + sin( x ) sin( y ) sin(2 x ) = 2 sin( x ) cos( x ) cos(2 x ) = sin
2
( x ) − cos
2
( x ) = 1 − 2 cos
2
( x ) = 2 sin
2
( x ) − 1
Exponent and Log Rules.
Assume b, c > 0 c x + y
= c x c y
, log c
( xy ) = log c
( x ) + log c
( y ) c
− x
= 1 /c x
, log c
(1 /x ) = − log c
( x )
( c x
) y
= c xy
, log c
( x y
) = y log c
( x ) b x c x
= ( bc ) x
, log b
( c ) = ln( c ) / ln( b )