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Find dy/dx in the following:
y=[x/√(1-x²)]- sin‾¹x

y'={[(1-x²)^(1/2)-(-2x)(x)(1/2)(1-x²)^(-1/2)]/(1-x²)} - (1/√(1-x²))

Then I multiplied (-2x)(x)(1/2) together. How do I simplify the rest?

  • Calculus -

    I agree so far with
    y' = {[(1-x²)^(1/2)-(-2x)(x)(1/2)(1-x²)^(-1/2)]/(1-x²)} - (1/√(1-x²))
    = {[(1-x²)^(1/2)-(-x^2(1-x²)^(-1/2)]/(1-x²)} - (1-x^2)^(1/2)/(1-x²)
    I formed a common denominator for the last term

    = (1-x^2)^(-1/2) [1-x^2 - x^2 - (1-x^2)]
    = -(1+x^2)/(1-x^2)^(3/2)

    There would be other variations, the trick is to recognize if they are the same

  • Calculus -

    Thank you for the help!

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