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#### DeusAbscondus

##### Active member

- Jun 30, 2012

- 176

for example:

$$y=ln(sinx)\Rightarrow y'=cot(x)$$ via the chain rule, including the following steps

$$\frac{d}{dx}ln(sinx)$$

$$=cos(x).\frac{1}{sin(x)}$$

$$=\frac{cos(x)}{sin(x)}$$

$=cot(x)$

but I would have expected that, following the chain rule, an operation was required on the $sin(x)$, but it just gets shunted down into the denominator along with $x$ Could someone kindly explain to me why?

I mean I understand this expression: $$y=ln(x)\Rightarrow y'=\frac {1}{x}$$

but how does the sine function make it down into the denominator holus-bolus in my example above?

I'm sorry for the awkwardness of expression.

Can anyone see the difficulty I am having with this?

Thanks,

DeusAbs