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Thread: Trigonometry Derivative ('Show That' Question)

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    Trigonometry Derivative ('Show That' Question)

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

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

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

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

    From the table of standard integrals, we know that

    $\displaystyle \int \frac{1}{a^2+b^2}\,dx = \frac{1}{a}arctan(\frac{x}{a})$
    But $\displaystyle \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)

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