Today we're going to talk about how to calculate the derivative of an inverse trig function
in particular the function here that we've written for y and because we have two arc
cotangent functions involved in our function here we've gone ahead and written the formula
for the derivative of arc cotangent when we're talking about the formulas for any inverse
trig functions it's important to remember that the x that you have inside of your inverse
trig function here is going to be the same as the x in the formula over here on the right
hand side so what that means is for example in the second part of our original function
when we have one divided by x inside of our cotangent function that means that in our
formula we're going to put one divided by x in place of x here so let's go ahead and
take the derivative and see what this looks like so our derivative function y prime will
be equal to arc cotangent of x we have the exact formula for this nothing different we
have to do so that's just negative one divided by one plus x squared for arc cotangent of
one divided by x we'll do the same thing here we have negative one divided by one plus but
in place of x here we'll put what we have inside of this arc cotangent function which
is one divided by x so one divided by x and we'll square that the other thing we have
to remember is that whenever we have anything other than x inside of our cotangent function
we need to multiply by the derivative of the inside function so that's a chain rule application
right chain rule tells us that we take the derivative of our outside function first so
we take the derivative of cotangent ignoring what's on the inside so we just took the derivative
of the outside function here leaving the inside function completely alone but now we have
to multiply by the derivative of the inside function here so the derivative of one divided
by x is negative one over x squared and the way that we know that remember is that one
over x is the same as x to the negative one we can use power rule then and say that the
derivative of that is negative x to the negative two and when we convert that back to a fraction
moving x to the negative two to the denominator to make the exponent positive we get negative
one over x squared so we multiply by the derivative of the inside function now our derivative
is just a matter of simplifying what we have here as much as possible so in order to do
that we'll first simplify the second part of this function so we have a negative sign
and a negative sign so those will cancel one another and become positive signs here in
our denominator we have this x squared term so we'll say x squared times the quantity
that we have here one plus one over x squared so we're going to get one plus one over x
squared and when we simplify that denominator further we'll distribute the x squared across
these two terms x squared times one is just x squared x squared times one over x squared
we get the x squared terms to cancel and we're just left with one so notice now that we have
the same two fractions here we just got the denominators in reverse order but one plus
x squared is obviously the same as x squared plus we have negative term and a positive
term which mean that we're actually going to get a result here of zero so that means
that our derivative is y prime is equal to zero because this negative fraction here cancels
with its opposite positive term these two go away and we're left with zero so I hope
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