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Math Help - Derivative Of Inverse Trigonometry, Exponents, And Log

  1. #1
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    Derivative Of Inverse Trigonometry, Exponents, And Log

    #1 Find y' of y=9^(-x)

    My work:
    y' = 9^(-x) * ln(9) * (-x)'
    y' = -9^(-x) * ln(9)


    #2 Find y' of y = ln((3-x)/(3+x))
    y' = 1/((3-x)/(3+x)) * ((3-x)/(3+x))'
    y' = 1/((3-x)/(3+x)) * ((3-x)'(3+x)-(3-x)(3+x)')/(3+x)^2
    y' = (3+x)/(3-x)* ((-1)(3+x)-(3-x)(1))/(3+x)^2
    y' = (3+x)/(3-x)* (-3-x-3+x)/(3+x)^2
    y' = 1/(3-x) * -6/(3+x)
    y' = -6/((x+3)(x-3)
    y' = -6/(x^2-9)

    #3 Find y' of y = ln((x(x^2+2)/(sqrt(x^3-7))


    Answers from my teacher:
    #1 (-2)(3^(-2x)(ln(3))
    #2 6/(x^2-9)
    #3 (1/x) + (2x)/(x^2+2) - (3x^2)/(2(x^3-7))
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  2. #2
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    For #1 notice that 9^{-x}=3^{-2x}.
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  3. #3
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    Quote Originally Posted by Plato View Post
    For #1 notice that 9^{-x}=3^{-2x}.
    Big help got #1 now
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  4. #4
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    At penultimate line of #2 there shouldn't be a minus sign since you change from (3-x) to -(x-3). Furthermore there is a much easier way of doing #2 by first applying the relationship of logarithms that states that

    \log_a \left( \frac{x}{y} \right) = \log_a x - \log_a y

    before differentiation.

    Trust me this way takes a lot less work!

    Applying the same method to #3 will also give you the required result quite easily.
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