# Why is the chain rule applied to derivatives of trigonometric functions?

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.

• Your notation is wrong - it should be written as $$\frac{d}{dx}(\cos 2x)$$ no $y$. – Thomas Andrews May 27 '15 at 22:56
• Thats good to know, Thanks Thomas - the notations are all so confusing... I rather use f'(x) or f''(x)... – Tiago Duque May 28 '15 at 11:56
• Rewrite $\cos2x$ as $\cos^2x-\sin^2x$ and you can clearly see that you can derivate it so easily. – Renato Faraone May 28 '15 at 12:09
• 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. – 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.