Suppose f(pi/6) = 7 and f’(pi/6) = 2

and let g(x) = f(x)cosx and h(x) = sinx/f(x)
g’(pi/6)=
h’(pi/6)=

1 answer

g' = f' cosx - f sinx
h' = (f' sinx - f cosx)/f^2
so plug in your numbers. Post your work if you get stuck
Similar Questions
  1. Simplify #3:[cosx-sin(90-x)sinx]/[cosx-cos(180-x)tanx] = [cosx-(sin90cosx-cos90sinx)sinx]/[cosx-(cos180cosx+sinx180sinx)tanx] =
    1. answers icon 1 answer
  2. hey, i would really appreciate some help solving for x when:sin2x=cosx Use the identity sin 2A = 2sinAcosA so: sin 2x = cos x
    1. answers icon 0 answers
  3. I have a question relating to limits that I solvedlim(x-->0) (1-cosx)/2x^2 I multiplied the numerator and denominator by
    1. answers icon 1 answer
  4. Trigonometric IdentitiesProve: (tanx + secx -1)/(tanx - secx + 1)= tanx + secx My work so far: (sinx/cosx + 1/cosx +
    1. answers icon 0 answers
more similar questions