What is the derivative of #cot^2(x)#?

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15
Oct 13, 2017

ANSWER
#d/dx cot^2(x)= -2cot(x)csc^2(x)#

EXPLANATION

You would use the chain rule to solve this. To do that, you'll have to determine what the "outer" function is and what the "inner" function composed in the outer function is.

In this case, #cot(x)# is the "inner" function that is composed as part of the #cot^2(x)#. To look at it another way, let's denote #u=cot(x)# so that #u^2=cot^2(x)#. Do you notice how the composite function works here? The "outer" function of #u^2# squares the inner function of #u=cot(x)#. The outer function determined what happened to the inner function.

Don't let the # u# confuse you, it's just to show you how one function is a composite of the other. You don't even have to use it. Once you understand this, you can derive.

The chain rule is:

#F'(x)=f'(g(x))(g'(x))#

Or, in words:
the derivative of the outer function (with the inside function left alone!) times the derivative of the inner function.

1) The derivative of the outer function #u^2=cot^2(x)# (with the inside function left alone) is:
#d/dx u^2= 2u#

(I'm leaving the #u# in for now but you can sub in #u=cot(x)# if you want to while you're doing the steps. Remember that these are just steps, the actual derivative of the question is shown at the bottom)

2) The derivative of the inner function:
#d/dx cot (x)= d/dx 1/tan(x) =d/dx sin(x)/cos(x)#

Hang on! You have to do a quotient rule here, unless you've memorized the derivative of #cot(x)#
#d/dx cos(x)/sin(x)=(-sin^2(x)-cos^2x)/(sin^2(x))=-(sin^2(x)+cos^2x)/(sin^2(x))= -1/(sin^2(x)) = -csc^2(x)#

Combining the two steps through multiplication to get the derivative:
#d/dx cot^2(x)= -2cot(x)csc^2(x)#

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