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

Question

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}))}$

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

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

Answer 1

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

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

Answer 2

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 $|x|$.

Answer 3

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 $$\frac{dy}{dx}=\frac{3}{x.\sqrt{x^{2}-9}}.$$