This seems straightforward to me, but it was marked wrong on my test with no clarification and no ability to ask questions.
$y=\operatorname{arcsec}(3x)$. Calculate the derivative.
I take the secant of both sides.
$$\sec(y)=3x$$
Take the derivative of both sides with respect to $x$.
$$\sec(y)\tan(y)y'=3$$
Solving for $y'$ gives:
$$y'= \frac{3}{(3x)(\sqrt{9x^2-1})}$$
The $3$'s cancel.
$$y'= \frac{1}{x\sqrt{9x^2-1}}.$$
Marked wrong. hmmmm... Can you assist me in what I did wrong?
$\endgroup$ 02 Answers
$\begingroup$The inverse secant and inverse cosecant functions are tricky ones. In different textbooks you can find different formulas for their derivatives — depending on the range chosen in the definitions of these functions. For example, for the inverse secant, its derivative can be given as $$(\operatorname{arcsec}x)'=\frac{1}{x\sqrt{x^2-1}} \quad \text{or} \quad (\operatorname{arcsec}x)'=\frac{1}{|x|\sqrt{x^2-1}},$$ depending on whether $\operatorname{arcsec}$ is defined to have the range of $$\left[0,\frac{\pi}{2}\right)\cup\left[\pi,\frac{3\pi}{2}\right) \quad \text{or} \quad \left[0,\frac{\pi}{2}\right)\cup\left(\frac{\pi}{2},\pi\right],$$ respectively.
Your answer is consistent with the first of these derivative formulas, but not with the second. So probably your textbook uses the second convention, which is is why your answer was marked wrong.
From the point of view of the solution that you presented, the issue is with the step when you decided that $\color{red}{\tan y=\sqrt{9x^2-1}}$. What we do know is that $\color{blue}{\tan^2y=9x^2-1}$, which implies that $\color{blue}{\tan y=\pm\sqrt{9x^2-1}}$. Whether the sign is plus or minus depends on the quadrant where $y$ lies — which effectively takes us back to knowing how the arcsecant function was defined (in your particular texbook).
$\endgroup$ $\begingroup$Your answer is correct. The answer key may have been looking for some equivalent form such as $$ \sqrt{\frac{1}{9x^4-x^2}} $$
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