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Chain Rule Problem

  1. Oct 25, 2014 #1
    1. The problem statement, all variables and given/known data

    Let g(x) = f(sin(2x) f(cos x)), where f(0) = 2, f'(0) = 3, f(-1) = -1/3 , and f'(-1) = -1. Find the equation of the tangent line to the curve of y = g(x) at x = pi.

    2. The attempt at a solution

    Point of Tangent: (pi, 2)
    g(pi) = f(sin(2pi) f(cos pi)) = f(0 * f(-1)) = f(0) = 2

    g(x) = f(sin(2x) f(cos x))
    = f'(sin(2x)* f(cos x)) * ([2cos(x) f(cos x)] + [f'(cos x) * (-sin x) * (sin 2x])
    = f'(sin(2pi) * f(cos pi)) *([2cos(pi) f(cos pi)] + [f'(cos pi) * (-sin pi) * (sin 2pi])
    = f'(0* f(-1)) *([(-2) f(-1)] + [f'(-1) * (0) * (0)])
    = f'(0) *(-2) f(-1)
    = (3)(-2)(-1/3) = 2

    Tangent line:

    y = 2x + b
    2 = 2(pi) + b
    b = 2 - (2)(pi)

    y = 2x + 2 - 2pi
     
  2. jcsd
  3. Oct 26, 2014 #2

    LCKurtz

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    Science Advisor
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    Isn't f(-1) = -1/3?

    You need to fix the derivative.
     
  4. Oct 26, 2014 #3
    g(pi) = f[ sin(2pi) * f (cos pi) ] = f [ 0 * -1/3 ] = f(0) = 2

    What's wrong with the derivative? I applied the chain rule.
     
  5. Oct 26, 2014 #4

    Doc Al

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    Staff: Mentor

    Looks like you messed up the derivative of sin(2x).
     
  6. Oct 26, 2014 #5
    Oh! It should be 2cos(2x). Thank you!
     
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