# How do you find the derivative of  y=arccos(1/x)?

Aug 8, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\left\mid x \right\mid \sqrt{{x}^{2} - 1}}$

#### Explanation:

There are a variety of methods that can be taken:

Method 1: Using the chain rule, knowing the arccosine derivative:

Know that the derivative of $\arccos \left(x\right)$ is $\frac{- 1}{\sqrt{1 - {x}^{2}}}$, so the derivative of $\arccos \left(f \left(x\right)\right)$ is $\frac{- 1}{\sqrt{1 - {\left(f \left(x\right)\right)}^{2}}} \cdot f ' \left(x\right)$.

Thus, we see that:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 1}{\sqrt{1 - {\left(\frac{1}{x}\right)}^{2}}} \cdot \frac{d}{\mathrm{dx}} \left(\frac{1}{x}\right) = \frac{- 1}{\sqrt{1 - \frac{1}{x} ^ 2}} \cdot \left(- \frac{1}{x} ^ 2\right)$

Continuing simplification:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{\frac{{x}^{2} - 1}{x} ^ 2}} \cdot \frac{1}{x} ^ 2 = \frac{1}{\frac{1}{\left\mid x \right\mid} \sqrt{{x}^{2} - 1}} \cdot \frac{1}{x} ^ 2 = \frac{1}{\sqrt{{x}^{2} - 1}} \cdot \frac{\left\mid x \right\mid}{x} ^ 2$

Here, note that $\frac{\left\mid x \right\mid}{x} ^ 2 = \frac{1}{\left\mid x \right\mid}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\left\mid x \right\mid \sqrt{{x}^{2} - 1}}$

Method 2: Rewriting first:

$y = \arccos \left(\frac{1}{x}\right) \text{ "=>" } \cos \left(y\right) = \frac{1}{x}$

Since cosine and secant are inverses:

$\sec \left(y\right) = x \text{ "=>" "y="arcsec} \left(x\right)$

You may already know the arcsecant derivative:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\left\mid x \right\mid \sqrt{{x}^{2} - 1}}$

Footnote to Method 2: Finding the Arcsecant Derivative

If $y = \text{arcsec} \left(x\right)$, how can we differentiate this? Start by rewriting:

$\sec \left(y\right) = x$

Now differentiate both sides. This is implicit differentiation and we'll need to use the chain rule on the left hand side. Remember that the derivative of $\sec \left(x\right)$ is $\sec \left(x\right) \tan \left(x\right)$.

$\sec \left(y\right) \tan \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

Then the derivative is:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sec \left(y\right) \tan \left(y\right)}$

First, let's think about this. Remember the original function is $y = \text{arcsec} \left(x\right)$, whose range is the same as the $\arccos \left(x\right)$ function: $y$ ranges from $0$ to $\pi$, meaning it only yields angles in the first and second quadrants.

Note that in the first quadrant, $\sec \left(y\right) \tan \left(y\right)$ is positive because both $\sec \left(y\right)$ and $\tan \left(y\right)$ are positive.

In the second quadrant, the product $\sec \left(y\right) \tan \left(y\right)$ is still positive because both $\sec \left(y\right)$ and $\tan \left(y\right)$ are negative.

Thus, considering the range of the original function, we see that $\sec \left(y\right) \tan \left(y\right)$ must be positive. This will be important in a moment.

Note that in the expression we found for $\frac{\mathrm{dy}}{\mathrm{dx}}$, we can rewrite $\tan \left(y\right)$ in terms of $\sec \left(y\right)$ via the Pythagorean identity:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sec \left(y\right) \sqrt{{\sec}^{2} \left(y\right) - 1}}$

Remember that $\sec \left(y\right) = x$:

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

But then, remember: $\sec \left(y\right) \tan \left(y\right) = x \sqrt{{x}^{2} - 1}$ must be positive. Note that $\sqrt{{x}^{2} - 1}$ always yields a positive result, so we add absolute value bars around $x$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\left\mid x \right\mid \sqrt{{x}^{2} - 1}}$