Question

Find the Derivative of F when: F(x)= (f(x)+2x)/(f(x)-2x)

Answer 1

since you are not given f, your answer must have an f' in it correct?

Answer 2

yes

Answer 3

then you need the quotient rule for this.
\[(\frac{h}{k})'=\frac{kh'-hk'}{k^2}\] with
\[h(x)=f(x)+2x, h'(x)=f'(x)+2, k(x)=f(x)-2x, k'(x)=f'(x)-2\]

Answer 4

so your denominator will be
\[(f(x)-2x)^2\]

Answer 5

numerator will be
\[(f(x)-2)\times (f'(x)+2)-(f(x)+2)\times (f'(x)-2)\] then algebra to clean it up

Answer 6

with quotient rule i get (f(x)-2x)(f'x+2)-(f(x)+2x))(f'x-2))/(f(x)-2))^2 but it want it simplified and im having trouble

Answer 7

make your life easy and replace
\[f(x)=y, f'(x)=y'\] should make algebra look simpler

Answer 8

alright thanks again!

Answer 9

for my answer choices its factoring out either a 2 or a 4 in the numerator

Answer 10

\[(y-2x)(y'+2)-(y+2x)(y'-2)\]
\[yy'+2y-2xy'-4x-(yy'-2y+2xy'-4x)\]

Answer 11

\[yy'+2y-2xy'-4x-yy'+2y-2xy'+4x\]

Answer 12

looks like we are left with nothing but
\[4y-4xy'\]

Answer 13

or if you prefer
\[4(y-xy)'\] or even
\[4(f(x)-xf'(x))\]

Answer 14

ok i think i got it now

Answer 15

check my algebra because i am doing it as i type, so i might have made an error, but i think it is right

Answer 16

would you mind helping me with one last one?

Answer 17

go ahead, i will help if i can

Answer 18

find the derivative of:
y=(cosx-1)/(sinx)

Answer 19

this should be ok

Answer 20

I derived it but cant simplify it in terms of their answer choices

Answer 21

\[y'=\frac{\sin(x)\times (-\sin(x))-(\cos(x)-1)\times \cos(x))}{\sin^2(x)}\] is a start

Answer 22

now we work in the numerator and get
\[-\sin^2(x)-\cos^2(x)+\cos(x)\] if i am not mistaken

Answer 23

and since
\[\sin^2(x)+\cos^2(x)=1\] we have
\[-\sin^2(x)-\cos^2(x)=-1\] so it looks like we get
\[\frac{\cos(x)-1}{\sin^2(x)}\]

Answer 24

we can go further i think

Answer 25

I got up to that point but all of my answer choices have a 1 in the numerator

Answer 26

since
\[\sin^2(x)=1-\cos^2(x)=(1-\cos(x))(1+\cos(x))\]

Answer 27

so we have
\[\frac{\cos(x)-1}{(1-\cos(x))(1+\cos(x))}\] and
\[\frac{\cos(x)-1}{1-\cos(x)}=-1\] we get
\[\frac{-1}{1+\cos(x)}\] or maybe
\[-\frac{1}{1+\cos(x)}\]

Answer 28

Thank you so much, you were very helpful!

Answer 29

there are many ways you can write this
yw

Answer 30

my big problem is sometimes I don't recognize some of the trig identities and it makes it very difficult to simplify. Thank You.