Calculus

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How do i find the second derivative of f(x)= tan(1-x)^1/2

  • Calculus -

    f'(x) = (1/2)(tan(1-x))^(-1/2)*sec^2 (1-x)(-1)
    = -(1/2) [tan(1-x)]^(-1/2) [sec(1-x)]^2

    now we need the product rule, what a mess ...

    f''(x) = -(1/2) [tan(1-x)]^(-1/2) (2)(sec(1-x)) (sec(1-x))(tan(1-x))(-1) + (-1/2)[sec(1-x)]^2 (-1/2)tan(1-x))^(-3/2)(sec^2(1-x))(-1)

    check this carefully and see if you simplify by factoring. I can see all kinds of common factors, but find it hard to read my own typing, lol

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