Derivatives of Trigonometric Functions

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• Misc
  • Date for hour exam 1?
    • Could move to Tuesday, Oct. 7
    • Test will be limits, limit definition of derivative, basic differentiation rules, product & quotient, trig derivatives
    • Problem set re trig derivatives will be due Thursday (Oct. 2), to be graded by Wednesday Oct. 8
    • (Mostly) up to you
    • Decision: exam is Tuesday, Oct. 7
      • 4 - 6 short-answer questions, e.g., like problem sets
      • Whole class period
      • Open book, notes, computer; closed person
• Questions?
• Derivatives of Trigonometric Functions
  • Section 3.5
    • Derivatives of sin, cos, tan
    • d sinx / dx = cosx
    • d cosx / dx = -sinx
    • d tanx / dx = sec²x
  • Examples: find f′(x) if...
    • f(x) = sinx - 3cosx
      • f′(x) = cosx + 3sinx
    • f(x) = (tanx)( x - sinx )
      • f′(x) = sec²x(x - sinx) + (1 - cosx) tanx  (product rule)
      • = (1/cos²x)(x - sinx) + (1 - cosx) tanx
    • f(x) = (sinx) / (x + secx)
      • u = sinx
      • v = (x + secx)
      • f′(x) = ( v du/dx - u dv/dx ) / (x + secx)²
      • = ( (x + secx) cosx - sinx (1 + secx tanx) ) / (x + secx)²
      • = ( (x + 1/cosx) cosx - sinx (1 + (1/cosx) (sinx/cosx)) ) / (x + secx)²
      • = ( (x + 1/cosx) cosx - sinx (1 + (sinx/cos²x) ) ) / (x + secx)²
      • = ( (x + 1/cosx) cosx - sinx - (sin²x/cos²x) ) / (x + secx)²
      • = ( x cosx + 1 - sinx - (sin²x/cos²x) ) / (x + secx)²
      • = ( x cosx + 1 - sinx - tan²x ) / (x + secx)²
      • = ( x cosx + 1 - sinx - tan²x ) / (x² + 2xsecx + sec²x)
  • Higher-order derivatives
    • Find d²( sinx + cosx ) / dx²
      • = (d/dx)( d(sinx + cosx) / dx )
      • =d( cosx - sinx ) / dx
      • = -sinx - cosx
    • Do 2nd, 3rd, etc. derivatives of sinx ever become 0?
      • No, they cycle between cosx, -sinx, -cosx, and sinx
  • Antiderivatives
    • Find a possible f(x) if f′(x) = -sinx
    • f(x) = cosx + C
  • Problem set
• Next
  • Silo problem
    
    • How fast is the tip of the shadow moving
    • x = 30 / tanΘ
    • Θ = sun’s angle above horizon
    • dΘ/dt = 2π radians/day = π/12 radians / hour
    • Want dx/dt
    • So calcuate dx/dΘ
      • = 30 d (1/tanΘ )/dx
      • = 30 d( cotΘ )/dx
      • = -30 csc²Θ feet/radian
    • Now multiply by dΘ/dt in radians/hour to get feet/hour
      • dx/dt = -30 csc²Θ feet/radian × π/12 radians / hour
      • = -2.5π csc²Θ feet/hour
      • Finally, express Θ as a function of time to get everything in terms of time
      • = -2.5π csc² πt/12  (t = time in hours)
  • You’ve just discovered the chain rule
    • Read about it in section 3.6 for Thursday

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