Question

I'm having a really hard time factoring... problem is find (dy/dx) of ((tanx)/(1+tanx)).... I know the answer and i know how to differentiate but the factoring I can't figure it out.. more in the draw form

Answer 1

\[(tanx)/(1+tanx) =(1+tanx)(\sec^2x)-(tanx)(\sec^2x)/(1+tanx)^2\]

Answer 2

how do i factor that to get the answer

Answer 3

First, I'll address your question (about factoring) right now. \[(1+tanx)(\sec^2x)-(tanx)(\sec^2x)\]has (sec x)^2 in both terms. Factor that (sec x)^2 out. What's left?

Answer 4

\[\sec^2x(1+tanx-tanx)\]

Answer 5

And the latter factor simplifies to what?

Answer 6

wow this is new stuff these derivatives I was way over thinking it thank you

Answer 7

\[\sin^2x/(1+tanx)^2\]

Answer 8

I have trouble with your equation\[(tanx)/(1+tanx) =(1+tanx)(\sec^2x)-(tanx)(\sec^2x)/(1+tanx)^2\]
Why? because you don't show the derivative operator. It's mandatory that you show it. Secondly, you need to use parentheses in such a way so that you and others realize that the entire right side of your equation is being divided by (1+tan x)^2.

You might want to think about re-typing what you meant.
This may sound picky, but most instructors would deduct points for the omissions I've mentioned.

Many thanks for the medal; hope to work with you again!

Answer 9

\[\frac{ dy }{dx }=\frac{ \left( 1+\tan x \right)\sec ^2x-\tan x \sec ^2x }{\left( 1+\tan x \right)^2 }\]
\[=\frac{ \sec ^2x \left( 1+\tan x-\tan x \right) }{ \left( 1+\tan x \right)^2 }\]
\[=\frac{ \sec ^2x }{ \left( 1+\tan x \right)^2 }\]

Answer 10

Thanks, I was trying to do that ^^^, for some reason it didn't work lol but I do understand about the things i left out again thank you both

Answer 11

yw