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Posted by on Tuesday, January 22, 2008 at 4:26pm.

Derivative of Inverse Trigonometric Functions

f(x) = sin(arccos(4x))

What is f'(x)?

  • Calculus III - , Tuesday, January 22, 2008 at 4:47pm

    f(x)= sin(arccos(4x)) = sqrt(1 - 16x^2)
    (To prove that, draw yourself a triangle with cos A = 4x and figure out the sin of A)

    let u = 1 - 16x^2
    f(u) = sqrt u
    f'(x) = df/dx = df/du du/dx
    = (1/2)(u)^-1/2 * -32 x
    = -16 x/sqrt[1 - 16x^2]

  • Calculus III - , Tuesday, January 22, 2008 at 8:38pm

    Thanks. The next question I have is pretty similar.

    f(x) =cos(arcsin(2x))

    Find f'(x)

  • Calculus III - , Tuesday, January 22, 2008 at 9:17pm

    It is so similar that you should be able to do it the same way. Draw the triangle and you have sqrt(1-4x^2)
    f(x) = (1-4x^2)^.5
    f'(x) = .5 [(1-4x^2)^-.5] (-8x)
    = -4x/sqrt(1-4x^2)

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