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