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lgbasalloteBest ResponseYou've already chosen the best response.0
have you tried quotient rule?
 one year ago

ERoseMBest ResponseYou've already chosen the best response.1
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
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.1
Show us how far you got with the quotient rule. Show all the steps you have so far. Can you do that?
 one year ago

ERoseMBest ResponseYou've already chosen the best response.1
Yes, I get (dw:1350172024457:dw
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.1
Correct! So far! Now what do you think the next step should be to simplify?
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.1
You're only going to simplify the numerator. Never expand the denominator. Do you understand?
 one year ago

ERoseMBest ResponseYou've already chosen the best response.1
Yes. Should I distribute the sinx to (1+secx)?
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.1
Go ahead and apply the distributive property to the numerator. In other words, expand.
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.1
Yes! Go ahead. Show me the next step.
 one year ago

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

calculusfunctionsBest ResponseYou've already chosen the best response.1
Yes expand the entire numerator. Meaning I shouldn't see any more parentheses in the numerator. Go ahead!
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.1
After that change each sec x to 1/cos x. Understood? Go ahead.
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.1
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?
 one year ago

ERoseMBest ResponseYou've already chosen the best response.1
Yes. I am just confused on what I should do to the cosx(tanxsecx)
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.1
\[\cos x(\tan x \sec x)=(\cos x)(\tan x)(\frac{ 1 }{ \cos x })\]Agreed?
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.1
So do exactly that. Go ahead.
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.1
Now show me the step in it's entirety.
 one year ago

calculusfunctionsBest 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.
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.1
You are correct in that you should change the sin x/cosx to tan x. Go ahead and show me the next step.
 one year ago

ERoseMBest ResponseYou've already chosen the best response.1
Can you multiply sinx and put it over cosx to get tangent on the left too?
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.1
Of course!\[\sin x(\frac{ 1 }{ \cos x })=\frac{ \sin x }{ \cos x }=\tan x\]
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.1
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!
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.1
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.
 one year ago

ERoseMBest ResponseYou've already chosen the best response.1
dw:1350174369510:dw Thank you for all your help
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.1
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.
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.1
The simplified answer should be\[f \prime(x)=\frac{ \sin x 2\tan x }{ (1+\sec x)^{2} }\]
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.1
So you understand where you went wrong?
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.1
Do you understand how to do this now?
 one year ago

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

ERoseMBest ResponseYou've already chosen the best response.1
Thank you so much for your help, thanks for following the process with me!
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.1
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.
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.1
My pleasure. If you'd like any more help, let me know.
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.1
Welcome anytime!
 one year ago
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