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Math Help - DERIVATE OF A INVERSE FUNCTION...

  1. #1
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    DERIVATE OF A INVERSE FUNCTION...

    y=\frac{1}{2x^2+3x-5} find y'

    could someone help me...no idea what to do...
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  2. #2
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    Quote Originally Posted by juanfe_zodiac View Post
    y=\frac{1}{2x^2+3x-5} find y'

    could someone help me...no idea what to do...
    Have you been taught the quotient rule? Use it here if you have.

    Alternatively, since y = (2x^2 + 3x - 5)^{-1} you can also use the chain rule. Have you been taught it?
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    Quote Originally Posted by juanfe_zodiac View Post
    y=\frac{1}{2x^2+3x-5} find y'

    could someone help me...no idea what to do...
    By the way, this is NOT an "inverse function", it is a reciprocal.
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    Have you been taught the quotient rule? Use it here if you have.

    Alternatively, since y = (2x^2 + 3x - 5)^{-1} you can also use the chain rule. Have you been taught it?

    from y = (2x^2 + 3x - 5)^{-1} i got

    (-x^{-2})(2x+3)(1) where x=(2x^2 + 3x - 5) for the first parenthesis.

    is it correct? i dont know hoe to continue it..help plz..i tried but i couldnt get it..tks
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  5. #5
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    Quote Originally Posted by juanfe_zodiac View Post
    from y = (2x^2 + 3x - 5)^{-1} i got

    (-x^{-2})(2x+3)(1) where x=(2x^2 + 3x - 5) for the first parenthesis.

    is it correct? Unfortunately No!
    i dont know hoe to continue it..help plz..i tried but i couldnt get it..tks
    f(x)=\dfrac1{2x^2+3x-5}=\left(2x^2+3x-5\right)^{-1}

    Let u(x) = 2x^2+3x-5 then f(x)=(u(x))^{-1}

    You have to use the chain rule to get the derivation:

    f'(x) = (-1)\cdot (u(x))^{-2}\cdot u'(x)

    With your question you have:

    f(x)=(-1)\cdot (2x^2+3x-5)^{-2} \cdot (4x+3)

    Re-write this equation without negative exponents:

    f(x)=-\dfrac{4x+3}{(2x^2+3x-5)^2}
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