Find f prime (x) if f(x) = (cosx)/(1+secx)

- anonymous

Find f prime (x) if f(x) = (cosx)/(1+secx)

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- katieb

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- lgbasallote

have you tried quotient rule?

- anonymous

Yes, I did and then I don't know how to distribute? I know I have to use it It just doesn't fully come out

- calculusfunctions

Show us how far you got with the quotient rule. Show all the steps you have so far. Can you do that?

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## More answers

- anonymous

Yes, I get (|dw:1350172024457:dw|

- calculusfunctions

Correct! So far! Now what do you think the next step should be to simplify?

- calculusfunctions

You're only going to simplify the numerator. Never expand the denominator. Do you understand?

- anonymous

Yes. Should I distribute the -sinx to (1+secx)?

- calculusfunctions

Go ahead and apply the distributive property to the numerator. In other words, expand.

- calculusfunctions

Yes! Go ahead. Show me the next step.

- anonymous

|dw:1350172306983:dw|
I shouldn't distribute the cosx correct? Because tanx and secx are being multiplied together?

- calculusfunctions

Yes expand the entire numerator. Meaning I shouldn't see any more parentheses in the numerator. Go ahead!

- calculusfunctions

After that change each sec x to 1/cos x. Understood? Go ahead.

- calculusfunctions

I mean change each sec x to 1/cos x only in the numerator. Never expand the denominator, like I said earlier. Do you understand?

- anonymous

Yes. I am just confused on what I should do to the -cosx(tanxsecx)

- calculusfunctions

\[-\cos x(\tan x \sec x)=-(\cos x)(\tan x)(\frac{ 1 }{ \cos x })\]Agreed?

- calculusfunctions

So do exactly that. Go ahead.

- calculusfunctions

Now show me the step in it's entirety.

- anonymous

Okay.

- anonymous

|dw:1350172978272:dw|

- calculusfunctions

\[f \prime(x)=\frac{ -\sin x(1+\sec x)-\cos x(\sec x \tan x) }{(1+\sec x)^{2} }\]
\[f \prime(x)=\frac{ -\sin x -\sin x \sec x -\cos x \sec x \tan x }{ (1+\sec x)^{2} }\]
\[f \prime(x)=\frac{ -\sin x -\sin x(\frac{ 1 }{ \cos x })-\cos x(\frac{ 1 }{ \cos x })\tan x }{ (1+\sec x)^{2} }\]
Take a moment to look at this the first couple of steps are exactly what you did. Do you understand what I did in the third step and now first tell me what you think you should do.

- calculusfunctions

You are correct in that you should change the sin x/cosx to tan x. Go ahead and show me the next step.

- anonymous

Can you multiply sinx and put it over cosx to get tangent on the left too?

- calculusfunctions

Of course!\[\sin x(\frac{ 1 }{ \cos x })=\frac{ \sin x }{ \cos x }=\tan x\]

- anonymous

|dw:1350174177943:dw|

- calculusfunctions

Have more confidence in yourself, would ya! You're doing just fine. Now continue with confidence. And even if you're wrong, so what? Being wrong is part of learning and part of life. So never be afraid to ask questions even if you think you're wrong. Now could you please show me the remaining steps, with confidence!

- calculusfunctions

Not quite. There is an error in the numerator but it just seems to be a careless one. Retrace your steps to see if you can spot the error. Otherwise show me your previous step so that I can point it out to you.

- anonymous

|dw:1350174369510:dw|
Thank you for all your help

- calculusfunctions

No! Now look back carefully at the third step of the partial solution, I posted above. Is sin x multiplied by tan x? I don't think so.

- calculusfunctions

The simplified answer should be\[f \prime(x)=\frac{ -\sin x -2\tan x }{ (1+\sec x)^{2} }\]

- anonymous

|dw:1350174836525:dw|

- calculusfunctions

So you understand where you went wrong?

- calculusfunctions

Do you understand how to do this now?

- anonymous

Yes, I think so, its just such a long process that I feel like I make mistakes so easily.

- anonymous

Thank you so much for your help, thanks for following the process with me!

- calculusfunctions

By the way, if the question doesn't ask you to simplify then it is not wrong to even leave your derivative unsimplified. However every teacher is different so you may want to consult with yours. I prefer students to simplify to a certain degree unless I specifically ask not to simplify. But as I said every teacher is different.

- calculusfunctions

My pleasure. If you'd like any more help, let me know.

- anonymous

Thank you so much!!

- calculusfunctions

Welcome anytime!

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