# Derivative of $\sec^{-1}(\frac{x}{3})$

Derivative of $\sec^{-1}(\frac{x}{3})$

I have tried these types of problems with two different approaches and keep getting the same answer which seems to be wrong. I suspect I am doing something obvious incorrectly; however, I can't seem to figure it out.

First method:

$y = \sec^{-1}(\frac{x}{3})$

therefore,

$\sec(y) = \frac{x}{3}$

$\frac{dy}{dx}\sec(y)\tan(y) = \frac{1}{3}$

$\frac{dy}{dx} = \frac{1}{3\sec(y)\tan(y)}$

Since $\sec(y) = \frac{x}{3}$ and $\tan(y) = \sqrt{\sec^2(y) -1} = \sqrt{\frac{x^2}{9} - 1}$

$\frac{dy}{dx} = \frac{1}{x\sqrt{\frac{x^2}{9} - 1}}$

Method two:

$(f')^{-1}(\frac{x}{3}) = \frac{1}{f'(\sec^{-1}(\frac{x}{3}))}$

$= \frac{1}{3\sec(\sec^{-1}{(\frac{x}{3})})\tan(\sec^{-1}(\frac{x}{3}))}$

$= \frac{1}{x\sqrt{\sec^2(\sec^{-1}(\frac{x}{3})) - 1}}$

$= \frac{1}{x\sqrt{\frac{x^2}{9} -1}}$

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WA says $\frac{d}{dx}\sec^{-1}(x/3)=3/\left(x^2\sqrt{1-\frac{9}{x^2}}\right)$, which confirms your results:

$$\begin{eqnarray} \frac{dy}{dx} &=& \frac{1}{x\sqrt{\frac{x^2}{9} - 1}} \text{ take out \frac{x^2}{9} from the \sqrt{\cdot} }\\ &=&\frac{1}{x\frac{x}{3}\sqrt{1-\frac{9}{x^2} }}\\ &=&\frac{3}{x^2\sqrt{1-\frac{9}{x^2} }} \end{eqnarray}$$

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How come when I integrate my answer I get: WA. Is it just a different form of $\sec^{-1}(\frac{x}{3})$? – stariz77 Jun 6 '12 at 11:26

There is something a little not quite right with your expression $$\frac{1}{x\sqrt{\frac{x^2}{9} -1}}$$ for the derivative of $\sec^{-1}(x/3)$. There are many numbers whose secant is $x/3$. So $\sec^{-1} u$ is defined as the number between $0$ and $\pi$ whose secant is $u$.

It is not hard to verify that this definition makes $\sec^{-1}$ an increasing function over any interval on which it is defined. Thus the derivative cannot be negative. However, when $x$ is negative your expression for the derivative is negative.

The fix is easy. Either use the Wolfram Alpha version, or in your version replace the $x$ outside the square root by $|x|$.

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Is there a way to arrive at the Wolfram Alpha version naturally, or does it require some observation like you have done after I have found the derivative, and then adjusting it accordingly? – stariz77 Jun 6 '12 at 12:43
Semi-naturally, since $9/x^2$ is $\cos^2 y$, so $\sqrt{1-9/x^2}$ is $\sin y$ (note that sine is positive over the interval). So it comes from computing in terms of sines and cosines. However, my inclination would be to compute crudely, being aware there could be sign problems, and fix at the end. – André Nicolas Jun 6 '12 at 12:53

Let $$y=\sec^{-1}\left(\frac{x}{3}\right)$$ $$\Rightarrow \sec y=\frac{x}{3}.$$ Differentiate with respect to $x$, we get, $$\sec y\tan y \frac{dy}{dx}=\frac{1}{3}$$ $$\therefore \sec y \sqrt{\sec^{2}y-1}\frac{dy}{dx}=\frac{1}{3}$$ $$\Rightarrow \frac{x}{3}\sqrt{\frac{x^{2}}{9}-1}\frac{dy}{dx}=\frac{1}{3}$$ $$\frac{x}{9}\sqrt{x^{2}-9}\frac{dy}{dx}=\frac{1}{3}$$ Thus $$\frac{dy}{dx}=\frac{3}{x.\sqrt{x^{2}-9}}.$$

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