1
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
2
$\begingroup$

$cos(2x)$ 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.