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

##### 1 Answer

#### 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

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

#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

#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(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

Note that in the first quadrant,

In the second quadrant, the product

Thus, considering the range of the original function, we see that **must** be positive. This will be important in a moment.

Note that in the expression we found for

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

Remember that

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

But then, remember: **must** be positive. Note that

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