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I'm stuck on differentiating this:

$$f(x) = \frac{4\sin(2x)}{e^\sqrt{2x-1}}$$

I thought about using the product rule here, but when I do that I get an expression that is hard to simplify, and I need to solve for when $f(x) = 0$.

Is there a simpler way of doing this?

Help here would be much appreciated!

Bob Christian
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2 Answers2

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Hint: if you want to use the product rule just write the function as

$$f(x) = 4\sin(x) e^{-\sqrt{2x-1}}$$

Then differentiate using the rule.

Spoiler

$$f'(x) = \frac{4 e^{-\sqrt{2 x-1}} \left(\sqrt{2 x-1} \cos (x)-\sin (x)\right)}{\sqrt{2 x-1}}$$

In order to have $f'(x) = $ you just need to solve

$$\sqrt{2x-1}\cos(x) - \sin(x) = 0$$

Can you proceed from here?

By the way are you sure the function you have is that one? For you get a transcendental equation...

Enrico M.
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  • Yes, the function is correct. It may not be relevant but the final part of the question which I excluded (I didnt think it was necessary to include it) was: 'The curve of y=f(x) has a maximum turning point at P and a minimum turning point at Q (from 0 to pi). Show that P and Q are the solutions to the equation tan2x=sqrt(2)'. Do I need to do something else? Or do I just solve f'(x)=0? – Bob Christian Mar 24 '20 at 15:23
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    It was 4sin(2x) sorry, but I guess now I can solve. Thank you for showing me the steps. – Bob Christian Mar 24 '20 at 15:27
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Instead of the product rule, I think that logarithmic differentiation would make life easier. $$f(x) = 4\sin(2x) e^{-\sqrt{2x-1}}\implies \log(f(x))=\log(4)+\log(\sin(2x))-\sqrt{2x-1}$$ $$\frac{f'(x)}{f(x)}=2\cot(2x)-\frac 1 {\sqrt{2x-1}}$$ Since you care about $f'(x)=0$, this is the equation to be solved.

Graphing you should notice a root "close" to $2.2$. Now, start Newton method.

Claude Leibovici
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