Expert: Ahmed Salami - 4/9/2009

Question
I was just wondering if you could help me with differentiating a few of these problems.  I tried to differentiate them to the best of my ability but I am not sure if I did them correctly.  Your help would be greatly appreciated!
1. h(x)=cos2x/sin2x
2. h(x)= [sin(1/x)][sin(1/x^2)]
3. If f(x)=(e^-x)(cosx), find the slope of the derivative at x=0. ****Side Note, e is referring to a constant, not a variable*****

Answer
Hi Nina,
1)h = cos2x/sin2x
Using the quotient rule for h = u/v
h' = (vu' - uv')/v^2       (i hope you understand all the shorthand)
u = cos2x
u' = -2sin2x
v = sin2x
v' = 2cos2x
h' = [(sin2x)(-2sin2x) - (cos2x)(2cos2x)] / (sin2x)^2
  = [-2(sin2x)^2 - 2(cos2x)^2] / (sin2x)^2
  = -2[(sin2x)^2 + (cos2x)^2] / (sin2x)^2
  = -2.1 / (sin2x)^2
Note: (sinA)^2 + (cosA)^2 = 1
  = -2/(sin2x)^2
  = -2(cosec2x)^2            (1/sinA = cosecA)
We could easily have done this by using the statement:
y = cotA
dy/dA = -(cosecA)^2
but its good you see the working.

2)h = [sin(1/x)][sin(1/x^2)]
Using the product rule for h = uv
h' = uv' - vu'
u = sin(1/x)
u' = -(1/x^2).cos(1/x)
v = sin(1/x^2)
v' = -(2/x^3).cos(1/x^2)
h' = [sin(1/x)][-(2/x^3).cos(1/x^2)] + [sin(1/x^2)][-(1/x^2).cos(1/x)]
  = -(2/x^3).sin(1/x).cos(1/x^2) - (1/x^2).sin(1/x^2).cos(1/x)

3)Yes, e is a constant, and not even an arbitrary constant, its the exponential constant.
Now, do you mean 'the slope OR the derivative'? Because its quite different from 'the slope OF the derivative', which is the slope of the slope i.e second differential. Anyway, i take it you mean the former, but be careful with words!
Proceeding as in question 2:
f(x) = (e^-x)(cosx)
u = e^-x
u' = -e^-x
v = cosx
v' = -sinx
f'(x) = (e^-x)(-sinx) + (cosx)(-e^-x)
     = -(e^-x)(sinx) - (e^-x)(cosx)
     = -(e^-x)(sinx + cosx)
at x = 0
f'(0) = -(e^0)(sin0 + cos0)
     = -1(0 + 1)
     = -1

Hope it helps you.

Regards