Ask your own question, for FREE!
Mathematics 83 Online
OpenStudy (anonymous):

can anyone help me on a quotient problem for Calculus on derivatives?

OpenStudy (anonymous):

whats the prob?

OpenStudy (anonymous):

y=(cosx)/(1-sinx) can you show the steps because I know I messed up somewhere but I just can extract exactly where.

OpenStudy (amistre64):

what have you tried?

OpenStudy (anonymous):

u know the quotient rule?

OpenStudy (amistre64):

\[[\frac{t}{b}]'=\frac{bt'-b't}{b^2}\]

OpenStudy (anonymous):

Yeah I do know the quotient rule.

OpenStudy (anonymous):

yes amistre, u definetly know this:)

OpenStudy (amistre64):

;)

OpenStudy (anonymous):

ok Dandy, apply it for ur prb

OpenStudy (anonymous):

Hmmm remember when you solve trig problems in calculus you have to simplify them using trig identity like Pythagorean theorem and double angel formula etc before you can eve get the correct answer

OpenStudy (anonymous):

Well right now I'm at (-sinx+sin^2x-cosx+cos^2x)/(-sinx)^2

OpenStudy (anonymous):

and I tried simplifying witht he trig identities but I'm not sure if I am applying them accordingly

OpenStudy (anonymous):

y'=(1-sinx)(-sinx)-(cosx)(-cosx)/(1-sinx)^2

OpenStudy (anonymous):

ohhhhhhhhhhhhhhhhh I see what I did wrong

OpenStudy (anonymous):

=(-sinx+sin^2x+cos^2)/(1-sinx)^2=-sinx+1/(1-sinx)^2

OpenStudy (anonymous):

I did the derivative wrong for v, it should of been -cosx because 1 is a constant.

OpenStudy (anonymous):

yes :)

OpenStudy (anonymous):

okay I think i did another computation error. now I'm at the final stages and I have (-sinx+1)/(1-sinx)^2

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!