Derivative of $\ln(1+\sin 2x)$ 
Differentiate the equation $y=\ln(1+\sin2x)$. 

It will be something to do with the $\frac{d}{dx}\{\ln\:x\}=\frac{1}{x}$ rule, but I'm not sure how to deal with the $\sin2x$ term.
 A: This has to do with the Chain Rule. Note that, without the application of the chain rule, you can just blindly use the derivative of $ \ln x $ and substitute the argument to get $$ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac {1}{1 + \sin 2x}, $$ but the chain rule says that you must multiply by the derivative of $ 1 + \sin 2x $, which is $ 2 \cos 2x $, which happens to be another application of the chain rule! Thus, the answer is $$ \frac {2 \cos 2x}{1 + \sin 2x}. $$Note that, in general, the chain rule says that the derivative of $ f(g(x)) $ is $ g'(x) \cdot f'(g(x)) $. 
A: Hint:
$$y(x) = \log{\square} \implies \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\square} \frac{\mathrm{d}\square}{\mathrm{d}x},$$
where $\square$ is anything depending on $x$.
A: The answer is $y'(x)=\frac{2\cos(2x)}{1+\sin(2x)}$. You are right about the rule. Everything inside $\ln$ must go in the denominator. The next step is to use the chain rule which says you multiply by the derivative of the interior argument of $\ln(1+\sin(2x))$ (i.e. multiply by the derivative of $1+\sin(2x)$)
A: Let $y=\ln(u); \quad u=1+\sin(v); \quad v=2x$
Then


*

*$$\frac{\mathrm{d}y}{\mathrm{d}u}=\frac{1}{u}=\frac{1}{1+\sin(v)}=\frac{1}{1+\sin(2x)} \quad ;$$

*$$\frac{\mathrm{d}u}{\mathrm{d}v}=\cos(v)=\cos(2x) \quad ;$$

*$$\frac{\mathrm{d}v}{\mathrm{d}x}=2 \quad .$$


Now use the fact that  $$\Large \boxed{\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}u} \cdot \frac{\mathrm{d}u}{\mathrm{d}v} \cdot \frac{\mathrm{d}v}{\mathrm{d}x} }\quad .$$

The boxed formula is an extension of the chain rule.
A: $y=\ln(1+\sin2x)$ can be written as $e^y=1+\sin2x$.
Differentiate this with respect to $x$ to get $e^y \cdot \frac{dy}{dx} = 2\cos2x$,
i.e. $\frac{dy}{dx} = 2\cos2x \cdot e^{(-y)}$
i.e. $\frac{dy}{dx} = \frac{2\cos2x}{(1+\sin2x)}$ :)
