Question

y=(2+secx)/(2-secx) find derivative??

Answer 1

make your life easy and rewrite first using cosines

Answer 2

\[\frac{2+\frac{1}{a}}{2-\frac{1}{a}}=\frac{2a+1}{2a-1}\] so what you really have here is
\[y=\frac{2\cos(x)+1}{2\cos(x)-1}\]

Answer 3

now you can take the derivative using the quotient rule, and you don't have to take the derivative of secant
you good from here or you still need the steps?

Answer 4

could you show me a few more steps please?

Answer 5

ok so lets take the derivative. it will be annoying because of cosine (not as annoying as it will be with secant) so lets first take the derivative of
\[y=\frac{2x+1}{2x-1}\] which is an easy quotient rule problem
you get
\[y'=\frac{-4}{(2x-1)^2}\] replacing x by cos(x) and using the chain rule you get
\[y'=\frac{4\sin(x)}{(2\cos(x)-1)^2}\]

Answer 6

if my replacements are annoying, you can find the derivative of
\[\frac{2\cos(x)+1}{2\cos(x)-1}\] using the quotient rule

Answer 7

\[\frac{(2\cos(x)-1)\times (-2\sin(x))-(2\cos(x)+1)\times (-2\sin(x))}{(2\cos(x)-1)^2}\]

Answer 8

then you will have a bunch of algebra to do in the numerator, and you will still get
\[y'=\frac{4\sin(x)}{(2\cos(x)-1)^2}\]

Answer 9

i think i understood it better when you changed it to cosine. thank so much!

Answer 10

yes, it is easier. but it take the sophistication to know that if i want the derivative of
\[\frac{2f(x)+1}{2f(x)-1}\] i might as well find the derivative of
\[\frac{2x+1}{2x-1}\] and then replace x by f(x) and multiply the result by f'(x) by the chain rule

Answer 11

how would you do this one when it has arcsin in it? y= 3x^4arcsin x

Answer 12

but it does not take much sophistacation to replace secant by 1/cos(x)and simplify,
if you do not do this you are goiong to get a huge mess

Answer 13

look in your book and find the derivative of arcsin(x) then use the product rule

Answer 14

\[\frac{d}{dx}\sin^{-1}(x)=\frac{}{\sqrt{1-x^2}}\] now use
\[(fg)'=f'g+g'f\]

Answer 15

\[\frac{d}{dx}\sin^{-1}(x)=\frac{1}{\sqrt{1-x^2}}\]

Answer 16

oh ok. gotcha! thanks