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Help with HW Question

Started by mikey4020, Oct 21 2007 02:05 PM

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#1 mikey4020

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Posted 21 October 2007 - 02:05 PM

I wonder if anyone can give me a hand with this question, it's been sticking me for a while now.

Show that $\begin{align*}f'(x)=2\sec x\end{align*}$ for $\begin{align*}f(x)=\ln (\frac{1+\sin x}{1-\sin x})\end{align*}$

thanks

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#2 George

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Posted 21 October 2007 - 04:22 PM

There are three important things here; $\frac{d}{dx}(\ln x)=\frac1x$ , the chain rule, and the quotient rule.

Using the first two things, note that if we have $f(x)=\ln g(x)$ , then $f'(x)=\frac{1}{g(x)}\times g'(x)$ . So in this case, we have

$\begin{align*}f'(x)=\frac{1}{\frac{1+\sin x}{1-\sin x}}\times\frac{d}{dx}\left(\frac{1+\sin x}{1-\sin x}\right)\end{align*}$

The first bit there is just $\begin{align*}\frac{1-\sin x}{1+\sin x}\end{align*}$ , and to get the second bit we use the quotient rule:

$\begin{align*}\frac{d}{dx}\left(\frac{1+\sin x}{1-\sin x}\right)&=\frac{\cos x(1-\sin x)-(1+\sin x)(-\cos x)}{(1-\sin x)^2}\\ &= \frac{\cos x(1-\sin x + 1 + \sin x)}{(1-\sin x)^2}\\ &= \frac{2\cos x}{(1-\sin x)^2}\end{align*}$

So, we now have

$\begin{align*}f'(x)=\frac{1-\sin x}{1+\sin x}\times\frac{2\cos x}{(1-\sin x)^2}\end{align*}$

and to get the result, you just need to cancel a $(1-\sin x)$ from top and bottom; then multiply out the denominator and use the fact that $1-\sin^2 x=\cos^2 x$ .

Do you get it now? Just let me know if any of that was unclear

Higher Maths Past Paper Solutions

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#3 mikey4020

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Posted 22 October 2007 - 11:25 AM

Thank, can't believe I never got that, it's fairly simple if you spot the important things at the start!!

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