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Mathematics Dictionary
Dr. K. G. Shih
Derivative : Trigonometric functions
Subjects


  • CA 03 00 | - Outines
  • CA 03 01 | - Derivative of y = sin(x)
  • CA 03 02 | - Derivative of y = cos(x)
  • CA 03 03 | - Derivative of y = tan(x)
  • CA 03 04 | - Derivative of y = csc(x)
  • CA 03 05 | - Derivative of y = sec(x)
  • CA 03 06 | - Derivative of y = cot(x)
  • CA 03 07 | - y = sin(n*x) find y'
  • CA 03 08 | - y = sin(x)^n find y'
  • CA 03 09 | - Find higher derivative of y = sin(x)
  • CA 03 10 | - Find higher derivative of y = cos(x)
  • CA 03 11 | - y = e^(sin(x)), find y'
    • Answers


    CA 03 01. Derivative of y = sin(x)

    Formula
    • y = sin(x) and y' = cos(x)
    Formula required to prove
    • Sin(A+B) = sin(A)*cos(B)+cos(A)*sin(B)
    • Lim[sin(x)/x] = 1 as x goes to zero
    Proof
    • y' = Lim[(sin(x+h) - sin(x))/h] when h tends to zero
    • y' = Lim[(sin(x)*cos(h) + cos(x)*sin(h) - sin(x))/h]
    • Since cos(h) = 1 when h = 0
    • Hence y' = Lim[cos(x)*sin(h)/h] = cos*x)*Lim[sin(h)/h]
    • Since Lim[sin(h)/h] = 1 as h goes to zero
    • Hence y' = cos(x)

    Go to Begin

    CA 03 02. Derivative of y = cos(x)

    Formula
    • y = cos(x) and y' = -sin(x)
    Formula required to prove
    • cos(A+B) = cos(A)*cos(B) - sin(A)*sin(B)
    • Lim[sin(x)/x] = 1 as x goes to zero
    Proof
    • y' = Lim[(cos(x+h) - cos(x))/h] when h tends to zero
    • y' = Lim[(cos(x)*cos(h) - sin(x)*sin(h) - cos(x))/h]
    • Since cos(h) = 1 when h = 0
    • Hence y' = Lim[-sin(x)*sin(h)/h] = -sin(x)*Lim[sin(h)/h]
    • Since Lim[sin(h)/h] = 1 as h goes to zero
    • Hence y' = -sin(x)
    Second method : Use cos(x) = sin(pi/2-x) and cos(x) = sin(pi/2-x)
    • Since cos(x) = sin(pi/2-x)
    • Take derivative on both sides
    • d/dx(cos(x)) = d/dx((sin(pi/2 - x))
    • d/dx(cos(x)) = sin(pi/2 - x)*d/dx(pi/2-x)
    • d/dx(cos(x)) = sin(pi/2 - x)*d/dx(pi/2-x)
    • Since d/dx(pi/2-x) = -1 and sin(pi/2-x) = cos(x)
    • Hence d/dx(cos(x)) = -sin(x)

    Go to Begin

    CA 03 03.Derivative of y = tan(x)

    Formula
    • y = tan(x) and y' = sec(x)^2
    Formula required to prove
    • tan(A+B) = (tan(x)+tan(y))/(1-tan(x)*tan(y)
    • Lim[tan(x)/x] = 1 as x goes to zero
    Proof
    • y' = Lim[(tan(x+h) - tan(x))/h] when h tends to zero
    • y' = Lim[((tan(x)+tan(h)/(1-tan(x)*tan(h) - tan(x))/h]
    • y' = Lim[((tan(x)+tan(h) - tan(x) + (tan(x)^2)*tan(h))/(h*(1-tan(x)*tan(h))]
    • Since tan(h) = 0 when h = 0
    • Hence y' = Lim[(1 + tan(x)^2)*(tan(h)/h]
    • Since Lim[tan(h)/h] = 1 as h goes to zero
    • Hence y' = 1 + tan(x)^2 = sec(x)^2
    Second method : Use tan(x) = sin(x)/cos(x) and quotient rule
    • Since tan(x) = sin(x)/cos(x)
    • Take derivative on both sides
    • d/dx(tan(x)) = d/dx((sin(x)/cos(x))
    • d/dx(cos(x)) = (d/dx(sin(x))*cos(x) - d/dx(cos(x))*sin(x))/(cos(x)^2)
    • d/dx(cos(x)) = (cos(x)^2 + sin(x)^2))/(cos(x)^2)
    • Since cos(x)^2 + sin(x)^2 = 1
    • Hence d/dx(tan(x)) = 1/(cos(x)^2) = sec(x)^2

    Go to Begin

    CA 03 04. Derivative of y = csc(x)

    Formula
    • d/dx(csc(x)) = -csc(x)*cot(x)
    Proof
    • Since csc(x) = 1/sin(x)
    • Use quotation rule
    • d/dx(csc(x)) = d/dx(1/sin(x))
    • d/dx(csc(x)) = (0)*sin(x)-(1)*cos(x))/(sin(x)^2)
    • d/dx(csc(x)) = (0)*sin(x)-(1)*cos(x))/(sin(x)^2)
    • d/dx(csc(x)) = -cos(x))/(sin(x)^2)
    • d/dx(csc(x)) = -csc(x)*cot(x)

    Go to Begin

    CA 03 05. Derivative of y = sec(x)
    Formula
    • d/dx(sec(x)) = sec(x)*tan(x)
    Proof
    • Since sec(x) = 1/cos(x)
    • Use quotation rule
    • d/dx(sec(x)) = d/dx(1/sin(x))
    • d/dx(sec(x)) = ((0)*sin(x)-(1)*cos(x))/(cos(x)^2)
    • d/dx(sec(x)) = ((0)*cos(x)-(1)*(-cos(x))/(cos(x)^2)
    • d/dx(sec(x)) = sin(x))/(cos(x)^2)
    • d/dx(sec(x)) = sec(x)*tan(x)

    Go to Begin

    CA 03 06.Derivative of y = cot(x)

    Formula
    • d/dx(cot(x)) = -csc(x)^2
    Proof
    • Since cot(x) = cos(x)/sin(x)
    • Use quotation rule
    • d/dx(cot(x)) = d/dx(cos(x)/sin(x))
    • d/dx(cot(x)) = (-sin(x))*sin(x)-(cos(x))*cos(x))/(sin(x)^2)
    • d/dx(cot(x)) = (-1*(sin(x)^2 + cos(x)^2))/(sin(x)^2)
    • d/dx(cot(x)) = -1)/(sin(x)^2)
    • d/dx(cot(x)) = -csc(x)^2

    Go to Begin

    CA 03 07. y = sin(n*x), find y'
    Solution
    • Let n*x = u and y = sin(u)
    • Then dy/dx = (d/du(sin(u))*(du/dx) = (cos(u))*(n) = n*cos(n*x)

    Go to Begin

    CA 03 08. y = sin(x)^n, find y'
    Solution
    • Let sin(x) = u and y = u^n
    • dy/dx = (d/du(u^n)*(du/dx)
    • dy/dx = (n*u^(n-1))*(cos(x))
    • dy/dx = n*(sin(x)^(n-1))*(cos(x))
    • dy/dx = n*cos(x)*((sin(x)^(n-1))

    Go to Begin

    CA 03 09. Find higher derivative of y = sin(x)

    nth derivative of sin(x)
    • 1st derivative of sin(x) = +cos(x)
    • 2nd derivative of sin(x) = -sin(x)
    • 3rd derivative of sin(x) = -cos(x)
    • 4th derivative of sin(x) = +sin(x)
    • 5th derivative of sin(x) = +cos(x)

    Go to Begin

    CA 03 10. Find higher derivative of y = cos(x)

    nth derivative of cos(x)
    • 1st derivative of cos(x) = -sin(x)
    • 2nd derivative of cos(x) = -cos(x)
    • 3rd derivative of cos(x) = +sin(x)
    • 4th derivative of cos(x) = +cos(x)
    • 5th derivative of cos(x) = -sin(x)

    Go to Begin

    CA 03 11. y = e^(sin(x)), find y'
    Solution
    • Take logarithm on both sides
    • We have ln(y) = (sin(x))*ln(e)
    • Since ln(e) = 1
    • Hence ln(y) = sin(x)
    • Derivative both sides with respective of x
    • d/dx(ln(Y)) = d/xd(sin(x))
    • (1/y)*(dy/dx) = cos(x)
    • dy/dx = y*cos(x)
    • dy/dx = (cos(x))*(e^(sin(x)))

    Go to Begin

    CA 03 00. Outlines

    Derivative of trigonometric functions
    • d/dx(sin(x)) = +cos(x)
    • d/dx(cos(x)) = -sin(x)
    • d/dx(tan(x)) = +sec(x)^2
    • d/dx(csc(x)) = -csc(x)*cot(x)
    • d/dx(sec(x)) = +sec(x)*tan(x)
    • d/dx(cot(x)) = -csc(x)^2
    Higher order derivative
    • (n+4)th derivative of sin(x) = sin(x) for n = 0, 1, 2, ...
    • (n+4)th derivative of cos(x) = cos(x) for n = 0, 1, 2, ...
    Derivative change curve of a function
    • Curve of y is sine curve
    • Curve of y' becomes cosine curve
    Curves of trigonometric functions
    Go to Begin
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