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Math Help - Trigonometry Derivative ('Show That' Question)

  1. #1
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    Trigonometry Derivative ('Show That' Question)

    Q: Given that y= \mathrm{arctan} \frac{x}{a}, where a is a constant, show that x \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\tan y}{1 + \tan ^2 y}.

    So... I have done (and reached a dead end):

    y= \mathrm{arctan} \frac{x}{a}
    \therefore \tan y = \frac {x}{a}
    \frac{\tan y}{1 + \tan ^2 y}
    = \frac{(\frac{x}{a})}{1 + (\frac{x}{a})^2}
    = \frac{(\frac{x}{a})}{\frac{a^2 + x^2}{a^2}}
    =\frac{ax}{a^2 + x^2}

    But That doesn't lead me anywhere. Please help. Thanks in advance.
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  2. #2
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    y = arctan(\frac{x}{a})

    From the table of standard integrals, we know that

    \int \frac{1}{a^2+b^2}\,dx = \frac{1}{a}arctan(\frac{x}{a})
    But \frac{1}{a}= 1 in our equation. Therefore a = 1 (not too sure about this step. I think if you went forward and worked with x/a it will cancel out anyway)

    y = arctan(\frac{x}{a})
     \frac{dy}{dx} = \frac{1}{1^2+ x^2}
    Multiplying both sides by x
     x\frac{dy}{dx} = \frac{x}{1+x^2}
    But x = tan (y)
    x\frac{dy}{dx} = \frac{tan (y)}{1+ tan^2(y)}
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