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

1 Answer
mason m
Aug 8, 2016

#dy/dx=1/(absxsqrt(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(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))#

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