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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.

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    $\begingroup$ Are you familiar with chain rule? $\endgroup$ – Dmoreno Aug 27 '14 at 20:56
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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)) $.

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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 $x$.

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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 $1+\sin(2x)$)

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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.

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$y=\ln(1+\sin2x)$ 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. $\frac{dy}{dx} = \frac{2\cos2x}{(1+\sin2x)}$ :)

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  • $\begingroup$ implicit differentiation? a few lessons ahead, don't you think? $\endgroup$ – BCLC Aug 27 '14 at 21:12
  • $\begingroup$ I'm not sure. I'm in India. I learnt implicit differentiation before Chain Rule... $\endgroup$ – Diya Aug 27 '14 at 21:14

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