Help with derivative of logs. Problem in comments

Question

anonymous · Answer

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anonymous · Answer

Hi Katie, how far have you got with this problem? Do you know where to begin?

anonymous · Answer

I haven't really started it. I'm confused on where to begin but I'm pretty sure you have to put it in log form.

anonymous · Answer

You can rearrange it using the fact that f(x)=e^ln(f(x) and that ln(a^b)=b*ln(a)

anonymous · Answer

So our equation becomes $$y=e^{{ln(x^{4\cos(x)}})}$$

anonymous · Answer

Now using the log rule I mentioned earlier, this becomes:
$$y=e^{4\cos(x)\ln(x)}$$

anonymous · Answer

And this is fairly easy to differentiate using the product rule :) Giving our answer:

$$ (-4sin(x)ln(x)+\frac{4cos(x)}{x})e^{4cos(x)ln(x)} $$

anonymous · Answer

Okay seems easy enough just a lot of work and not knowing where to start. Thank you :)

anonymous · Answer

No problem, glad you understand it. Yeah it can be a bit tricky to spot how to do this, but the f(x)=e^ln(f(x)) trick is very useful in many branches of maths.

bibby · Answer

$ln(a^b)=bln(a)$
wouldn't it be easier to rewrite it as $lny=4cos(x)ln(x)$
$\frac{1}{y}y'=\frac{d}{dx}4cosxlnx$?

anonymous · Answer

I wouldn't say it's any easier (or harder for that matter), but yes it's another approach. Knowing how to use e^ln is very useful though, which is why I wanted to do it that way.