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 MATH 150 Calculus I ٍ

Sample Test

1 a) [5] f(x) = (3 x2 + sqrt(x) + 1)10. What is f '(x)?

Solution:1. 10 (3 x2 + 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) = t49(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 x2 + xy + y2 = 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

 

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