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Help with HW Question - HSN forum

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Help with HW Question


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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 smile.gif

#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 smile.gif

#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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