I'm having trouble to understand why is the Chain rule applied to trigonometric functions, like: $$\frac{d}{dx}\cos 2x=[2x]'*[\cos 2x]'=-2 \sin 2x$$ Why isn't it like in other variable derivatives? Like in: $$ \frac{d}{dx} 3x^2=[3x^2]'=6x $$ Does it means it is the derivative of the trig function times the derivative of the angle?

Thanks once again.

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    $\begingroup$ Your notation is wrong - it should be written as $$\frac{d}{dx}(\cos 2x)$$ no $y$. $\endgroup$ – Thomas Andrews May 27 '15 at 22:56
  • $\begingroup$ Thats good to know, Thanks Thomas - the notations are all so confusing... I rather use f'(x) or f''(x)... $\endgroup$ – Tiago Duque May 28 '15 at 11:56
  • $\begingroup$ Rewrite $\cos2x$ as $\cos^2x-\sin^2x$ and you can clearly see that you can derivate it so easily. $\endgroup$ – Renato Faraone May 28 '15 at 12:09
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    $\begingroup$ A problem with the $f'$ notation is that you get expressions like $[\cos(2x)]'$ that seem ambiguous. Does that mean to take the derivative of $f(x) = \cos(x)$ and apply it to $2x$, or does it mean to take the derivative of $f(x) = \cos(2x)$ and apply it to $x$? Clearly (from the answer) you mean the first interpretation, but when you evaluate $[3x^2]'$ the answer looks more like the second interpretation. $\endgroup$ – David K May 28 '15 at 12:09

$cos(2x)$ is a chain of two functions

$f(x)=2x$ and $g(x)=cos(x)$

You have to calculate the derivate of $g(f(x))$ and for this, you need the chain rule.

The example $f(x)=3x^2$ can be derivated with the factor-rule and the power-rule. You need no chain-rule here.


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