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

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

EXPLANATION

You would use the to reresolve this. To do that, you'll enjoy to portioicularize what the "outer" operation is and what the "inner" operation collected in the outside operation is.

In this contingency, ##cot(x)## is the "inner" operation that is collected as portio of the ##cot^2(x)##. To observe at it another way, let's resemble ##u=cot(x)## so that ##u^2=cot^2(x)##. Do you regard how the composite operation works hither? The "outer" operation of ##u^2## squares the close operation of ##u=cot(x)##. The outside operation portioicularized what happened to the close operation.

Don't let the ## u## disorganize you, it's lawful to profession you how one operation is a composite of the other. You don't level enjoy to use it. Once you apprehend this, you can draw.

The fastening government is:

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

Or, in words: the derivative of the outside operation (delay the internally operation left singular!) times the derivative of the close operation.

1) The derivative of the outside operation ##u^2=cot^2(x)## (delay the internally operation left singular) is: ##d/dx u^2= 2u##

(I'm leaving the ##u## in for now but you can sub in ##u=cot(x)## if you absence to suitableness you're doing the steps. Remember that these are lawful steps, the objective derivative of the investigation is professionn at the depth)

2) The derivative of the close operation: ##d/dx cot (x)= d/dx 1/tan(x) =d/dx sin(x)/cos(x)##Hang on! You enjoy to do a hither, cosmical you've memorized the derivative of ##cot(x)## ##d/dx cos(x)/sin(x)=(-sin^2(x)-cos^2x)/(cos^2(x))=-(sin^2(x)+cos^2x)/(cos^2(x))=-csc^2(x)##

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

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