What is the Derivative of cos(sin(x)) and Why is it Different from -sin(sin(x))?

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SUMMARY

The derivative of cos(sin(x)) is correctly calculated as -sin(sin(x)) * cos(x) using the chain rule. The confusion arose from a misinterpretation of the differentiation process, where the derivative of the outer function cos(g) is -sin(g) evaluated at g = sin(x). This highlights the importance of correctly applying the chain rule in calculus to avoid common pitfalls in differentiation.

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monet A
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Hi I'm new here, great forum!
On another thread that I can't find I found a message from jamesrc about differentiating cos of sinx that says:

df(g(x))/dx = df(g)/dg * dg(x)/dx by the chain rule, this part I get and I thank james for the help the thing that confuses me is why df(g)/dg = cos(cosx).
Aren't we differentiating df(g) wrt (g) which would give -sin(sinx). What am I missing?
 
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You're not missing anything; you are correct: d/dx [\cos (\sin x)] = - \sin (\sin x) \cos x
 
Oh thanks Al! :smile:
I just found the thread I read and realized that james was differentiating sin of cos not cos of sin, lol.

:blushing:
 
I'm glad you cleared jamesrc's good name! :smile:

And welcome to PF, by the way.
 

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