I'm trying to find the derivative of this:

y= cot(cos(x))^2

what im getting is:

y'= -csc(cos(x))^4 * 2cos(x) * -sin(x)

but thats not working.

any ideas here?
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first and foremost
do you mean
y= cot^2(cosx)
or
y= cot(cos^2 (x))
the derivative of the first needs power rule, trig rules, and chain rule

in the first form it can be rewritten as

y=cot(cos(x))cot(cosx) then do chain rule and product rule, you should get it eventually, then again I might be wrong.
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to clarify its

y=cot(cos^2(x))
I'm getting 2(cot(cosx))(sinx)(csc(cosx))

Assuming that it's cotangent squared of cosine x.

EDIT: You bitch!

In that case, I think it's 2(csc^2(cosx))(cosx)(sinx)
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Last edited by abcdboy at Feb 24, 2009,
i think its y'=-csc^2[2(cosx)(-sinx)]... the derivative of cotx is (-cscx)^2, so its that of 2 (b/c thats the power) times the term in the parentheses (down a power) times the derivative of the term in the parentheses. thereyago.
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i think what you got is the same thing except the 2cosxsinx is inside becuase its what you get with the chain rule
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Quote by abcdboy

In that case, I think it's 2(csc^2(cosx))(cosx)(sinx)

your first 'cos x' should be '(cos x)^2.'

just to make it a bit clearer to the OP, it's chain rule inside a chain rule. first differentiate cot y [where y=cos^2(x)] to get -cosec^2(y) but then you need to differentiate y as well, as dictated by the chain rule.

y is technically a function of a function too, albeit a rather trivial one. the derivative of (cos x)^2 = 2(cos x) . -sin x, which you need to multiply by your -cosec^2 malarky. the minuses cancel and there you go.
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Last edited by Sol9989 at Feb 24, 2009,