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ERoseM
 3 years ago
Find f prime (x) if f(x) = (cosx)/(1+secx)
ERoseM
 3 years ago
Find f prime (x) if f(x) = (cosx)/(1+secx)

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lgbasallote
 3 years ago
Best ResponseYou've already chosen the best response.0have you tried quotient rule?

ERoseM
 3 years ago
Best ResponseYou've already chosen the best response.1Yes, 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
 3 years ago
Best ResponseYou've already chosen the best response.1Show us how far you got with the quotient rule. Show all the steps you have so far. Can you do that?

ERoseM
 3 years ago
Best ResponseYou've already chosen the best response.1Yes, I get (dw:1350172024457:dw

calculusfunctions
 3 years ago
Best ResponseYou've already chosen the best response.1Correct! So far! Now what do you think the next step should be to simplify?

calculusfunctions
 3 years ago
Best ResponseYou've already chosen the best response.1You're only going to simplify the numerator. Never expand the denominator. Do you understand?

ERoseM
 3 years ago
Best ResponseYou've already chosen the best response.1Yes. Should I distribute the sinx to (1+secx)?

calculusfunctions
 3 years ago
Best ResponseYou've already chosen the best response.1Go ahead and apply the distributive property to the numerator. In other words, expand.

calculusfunctions
 3 years ago
Best ResponseYou've already chosen the best response.1Yes! Go ahead. Show me the next step.

ERoseM
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1350172306983:dw I shouldn't distribute the cosx correct? Because tanx and secx are being multiplied together?

calculusfunctions
 3 years ago
Best ResponseYou've already chosen the best response.1Yes expand the entire numerator. Meaning I shouldn't see any more parentheses in the numerator. Go ahead!

calculusfunctions
 3 years ago
Best ResponseYou've already chosen the best response.1After that change each sec x to 1/cos x. Understood? Go ahead.

calculusfunctions
 3 years ago
Best ResponseYou've already chosen the best response.1I mean change each sec x to 1/cos x only in the numerator. Never expand the denominator, like I said earlier. Do you understand?

ERoseM
 3 years ago
Best ResponseYou've already chosen the best response.1Yes. I am just confused on what I should do to the cosx(tanxsecx)

calculusfunctions
 3 years ago
Best ResponseYou've already chosen the best response.1\[\cos x(\tan x \sec x)=(\cos x)(\tan x)(\frac{ 1 }{ \cos x })\]Agreed?

calculusfunctions
 3 years ago
Best ResponseYou've already chosen the best response.1So do exactly that. Go ahead.

calculusfunctions
 3 years ago
Best ResponseYou've already chosen the best response.1Now show me the step in it's entirety.

calculusfunctions
 3 years ago
Best ResponseYou've already chosen the best response.1\[f \prime(x)=\frac{ \sin x(1+\sec x)\cos x(\sec x \tan x) }{(1+\sec x)^{2} }\] \[f \prime(x)=\frac{ \sin x \sin x \sec x \cos x \sec x \tan x }{ (1+\sec x)^{2} }\] \[f \prime(x)=\frac{ \sin x \sin x(\frac{ 1 }{ \cos x })\cos x(\frac{ 1 }{ \cos x })\tan x }{ (1+\sec x)^{2} }\] 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
 3 years ago
Best ResponseYou've already chosen the best response.1You are correct in that you should change the sin x/cosx to tan x. Go ahead and show me the next step.

ERoseM
 3 years ago
Best ResponseYou've already chosen the best response.1Can you multiply sinx and put it over cosx to get tangent on the left too?

calculusfunctions
 3 years ago
Best ResponseYou've already chosen the best response.1Of course!\[\sin x(\frac{ 1 }{ \cos x })=\frac{ \sin x }{ \cos x }=\tan x\]

calculusfunctions
 3 years ago
Best ResponseYou've already chosen the best response.1Have 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
 3 years ago
Best ResponseYou've already chosen the best response.1Not 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.

ERoseM
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1350174369510:dw Thank you for all your help

calculusfunctions
 3 years ago
Best ResponseYou've already chosen the best response.1No! 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
 3 years ago
Best ResponseYou've already chosen the best response.1The simplified answer should be\[f \prime(x)=\frac{ \sin x 2\tan x }{ (1+\sec x)^{2} }\]

calculusfunctions
 3 years ago
Best ResponseYou've already chosen the best response.1So you understand where you went wrong?

calculusfunctions
 3 years ago
Best ResponseYou've already chosen the best response.1Do you understand how to do this now?

ERoseM
 3 years ago
Best ResponseYou've already chosen the best response.1Yes, I think so, its just such a long process that I feel like I make mistakes so easily.

ERoseM
 3 years ago
Best ResponseYou've already chosen the best response.1Thank you so much for your help, thanks for following the process with me!

calculusfunctions
 3 years ago
Best ResponseYou've already chosen the best response.1By 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
 3 years ago
Best ResponseYou've already chosen the best response.1My pleasure. If you'd like any more help, let me know.

calculusfunctions
 3 years ago
Best ResponseYou've already chosen the best response.1Welcome anytime!
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