** MATH 150 Calculus I**** ٍ**

**Sample Test**

**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**