BRIEF ANSWERS FOR TEST 1

1. T or F?
   1. F. [sec(x)]' = sec(x)*tan(x)
   2. F. [log_a(x)]' = 1/(x*ln(a))
   3. F. [f^(-1)(x)]' = 1/f'(f^(-1)(x)). To see that this is different, take f(x) = x^2.
   4. F. For example, f(x) = -abs(x) has an absolute maximum at x = 0, but it is not differentiable at x = 0.
2. Give the definition.
   1. f'(x) = lim_{h goes to 0} (f(x+h)-f(x))/h.
   2. f has a local max at x = c if f(c) is greater than or equal to f(x) for all x in an open interval around c.
   3. x = c is a critical value of y = f(x) if f'(c) = 0 or f'(c) does not exist.
3. Find the limits.
   1. Let f(x) = sqrt(x). The limit we want is f'(100) = 1/20.
   2. Let f(x) = 2^x. The limit we want is f'(0) = ln(2).
4. If f(x) = x*abs(x), then f'(0) exists and equals 0. To see this, apply the definition: lim_{h goes to zero} (h*abs(h))/h = lim_{h goes to zero} abs(h) = 0.
   1. f'(x) = sec^2(x) + 1/(1+x^2)
   2. g'(x) = x^x*(1+ln(x)) + 2^x*ln(2) + 2*x
   3. h'(x) = (e^(sin(ln(x)))*cos(ln(x)))/x
5. Implicit differentiation yields that dy/dx = (1+ln(x))/(1+ln(y)), so at (x,y) = (1/2,1/4) the derivative is (1+ln(1/2))/(1+ln(1/4)).
6. Let f(x) = sqrt(x). Then the linear approximation to f at x = 64 is L(x) = f(64) + f'(64)*(x-64) = 8 + (1/16)*(x-64). Thus L(65) = 8 + 1/16.
7. Let y = the height of the shadow. Let x = the distance between the man and the spotlight. An argument with similar triangles shows that x/2 = 12/y, so x*y = 24. Thus dy/dt = (-24/x^2)*(dx/dt). When x = 8, this gives dy/dt = (-24/8^2)*(1.6) = -0.6 m/s.