Pre-Calculus Test 2
Instructor: Prof. A. Mingarelli
Feb 20, 1997
NOTATION:
Pi = 3.14159... is our usual Greek number
1 a) [5] f(x) = (3 x
2
+ sqrt(x) + 1)
10
. What is f '(x)?
Solution:
1. 10 (3 x
2
+ sqrt(x) + 1)
9
(6x + 1/2sqrt(x)) by the Chain Rule.
1 b) [5] f(t) = |t-4|. What is f '(4)? Explain.
Solution:
f '(4) does not exist...Why? For t < 4, f(t) = 4-t and so its left derivative at t = 4 is -1. On the other hand for t > 4, f(t) = t - 4, so its right derivative at t = 4 is +1. Since these are different, the function cannot be differentiable there.
2. [10] Find points on the curve y = (sqrt(t))
100
/(1+t) where y '(t) = 0.
Solution:
Use the Quotient Rule: You'll get y '(t) = t
49
(50-t)/(1+t)
2
. Set this expression equal to zero and you'll find t = 0 or t = 50.
3. [5+5] Two differentiable functions f, g, are given with the following properties: f '(0) = -1, f(0) = 1, g(0) = 6, and g '(0) = -2. Evaluate (fg) '(0) and (f/g) '(0).
Solution:
(fg) '(0) = -8 and (f/g) '(0) = -1/9... Use the definition of the Product and Quotient Rules and set the variable = 0 therein.
4. [10] Given that x
2
+ xy + y
2
= 7 and y is a differentiable function of x, evaluate the derivative y ' at the point (x,y) = (2,1)
Solution:
Use implicit differentiation: Since y is a function of x we have 2x + xy ' + y + 2yy' = 0. Solving for y' we get y' = -(2x + y)/(x+2y). So, at (2,1) we find y' = -5/4 = -1.
Total: 40 points