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Thread: inverse function

  1. #1
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    inverse function

    f(x)=tan^-1 (-x/x+1)

    show that f'(x)=-1/2x^2+2x+1


    using the formula tan^-1 x/a =a/a^2+x^2


    f'(x)=-x/(x+1)^2 +(-x)^2

    sadly this is not right
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  2. #2
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    Re: inverse function

    Quote Originally Posted by markosheehan View Post
    f(x)=tan^-1 (-x/x+1)
    -x/x+ 1= 0 for all x. I am sure you mean tan^1(-x/(x+ 1)).

    show that f'(x)=-1/2x^2+2x+1
    and here you mean f'(x)= -1/(2x^2+ 2x+ 1), right?


    using the formula tan^-1 x/a =a/a^2+x^2
    You mean the derivative of the arctan of x/a is equal to that, not the arctan itself.
    The derivative of tan^1 x= 1/(1+ x^2) (Please use parentheses!) so first replace that "x" by "-x/(x+ 1)":
    1/(1+ x^2/(x+ 1)^2).
    Simplify that by multiplying numerator and denominator by (x+ 1)^2: (x+1)^1/((x+1)^2+ x^2= (x+ 1)^2/(2x^2+ 2x+ 1).

    Now, using the chain rule, multiply that by the derivative of -x/(x+ 1). Using the quotient rule, that is [-(x+ 1)- 1(-x)]/(x+ 1)^2= -1/(x+ 1)^2 to get
    [(x+ 1)^2/(2x^2+ 2x+ 1)][-1/(x+ 1)^2]= -1/(2x^2+ 2x+ 1).

    f'(x)=-x/(x+1)^2 +(-x)^2

    sadly this is not right
    Thanks from markosheehan
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  3. #3
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    Re: inverse function

    Quote Originally Posted by HallsofIvy View Post
    -x/x+ 1= 0 for all x. I am sure you mean tan^1(-x/(x+ 1)). . . . . . tan^-1(-x/(x+ 1))



    The derivative of tan^1 x= 1/(1+ x^2) . . . . . tan^-1(x) = 1/(1+ x^2)


    Simplify that by multiplying numerator and denominator by (x+ 1)^2:

    (x+1)^1/((x+1)^2+ x^2 . . . . . (x+1)^1/((x+1)^2+ x^2)


    = (x+ 1)^2/(2x^2+ 2x+ 1).
    .
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  4. #4
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    Re: inverse function

    yes you have read my post right.

    i understand you up to the point where you say
    "Now, using the chain rule, multiply that by the derivative of -x/(x+ 1). Using the quotient rule, that is [-(x+ 1)- 1(-x)]/(x+ 1)^2= -1/(x+ 1)^2 to get
    [(x+ 1)^2/(2x^2+ 2x+ 1)][-1/(x+ 1)^2]= -1/(2x^2+ 2x+ 1)."

    i do not see why we have to use the chain rule. the formula tan^1 x= 1/(1+ x^2) is supposed to give you the derivative it for you i thought. i do not see why you multiply by the derivative of -x/(x+ 1)
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  5. #5
    Junior Member Ahri's Avatar
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    Re: inverse function

    $y=\arctan(\frac{-x}{x+1})$

    Given $\frac{d}{dx} \arctan(\frac{x}{a})=\frac{a}{a^2+x^2}$, for $a=1$, $\frac{d}{dx}\arctan(x)=\frac{1}{1+x^2}$

    Let $u(x)= \frac{-x}{x+1}$, with $\frac{du}{dx}=\frac{-(x+1)+x}{(x+1)^2}=\frac{-1}{(x+1)^2}$ (Quotient rule).

    Now $y(u)=\arctan(u)$ and $\frac{dy}{du}=\frac{1}{1+u^2}$ which is given in the question.

    By the chain rule, $\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$

    Can you finish it from there?
    Last edited by Ahri; Jul 12th 2017 at 08:29 AM.
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