Question

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

Answer 1

`dy/dx=1/(absxsqrt(x^2-1))`

Explanation:

There are a variety of methods that can be taken:

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

Know that the derivative of `arccos(x)` is `(-1)/sqrt(1-x^2)`, so the derivative of `arccos(f(x))` is `(-1)/sqrt(1-(f(x))^2)*f'(x)`.

Thus, we see that:

`dy/dx=(-1)/sqrt(1-(1/x)^2) * d/dx(1/x)=(-1)/sqrt(1-1/x^2) * (-1/x^2)`

Continuing simplification:

`dy/dx=1/sqrt((x^2-1)/x^2)*1/x^2=1/(1/absxsqrt(x^2-1))*1/x^2=1/sqrt(x^2-1)*absx/x^2`

Here, note that `absx/x^2=1/absx`:

`dy/dx=1/(absxsqrt(x^2-1))`

Method 2: Rewriting first:

`y=arccos(1/x)" "=>" "cos(y)=1/x`

Since cosine and secant are inverses:

`sec(y)=x" "=>" "y="arcsec"(x)`

You may already know the arcsecant derivative:

`dy/dx=1/(absxsqrt(x^2-1))`

Footnote to Method 2: Finding the Arcsecant Derivative

If `y="arcsec"(x)`, how can we differentiate this? Start by rewriting:

`sec(y)=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(x)` is `sec(x)tan(x)`.

`sec(y)tan(y)dy/dx=1`

Then the derivative is:

`dy/dx=1/(sec(y)tan(y))`

First, let's think about this. Remember the original function is `y="arcsec"(x)`, whose range is the same as the `arccos(x)` 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(y)tan(y)` is positive because both `sec(y)` and `tan(y)` are positive.

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

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

Note that in the expression we found for `dy/dx`, we can rewrite `tan(y)` in terms of `sec(y)` via the Pythagorean identity:

`dy/dx=1/(sec(y)sqrt(sec^2(y)-1))`

Remember that `sec(y)=x`:

`dy/dx=1/(xsqrt(x^2-1))`

But then, remember: `sec(y)tan(y)=xsqrt(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`:

`dy/dx=1/(absxsqrt(x^2-1))`