Question

first derivative of

Answer 1

y= y arctan e ^-2x

Answer 2

Could you write this out more clearly?

Answer 3

\[y=y \arctan e ^{-2x}\]

Answer 4

@Kainui

Answer 5

I don't think that's right. You can divide both sides by y to get:

1=arctan(e^(-2x))

and here we can see that you can just solve for x, it's just a constant.

Answer 6

Unless that's the point, in which case the derivative of a constant is 0.

Answer 7

its actually a derivation of natural logarithm

Answer 8

@Kainui u can't divide both sides by y bcoz it's variable and may y=0

Answer 9

u can find it by implicit derivative

Answer 10

K, good luck guys, clearly I have no idea what's going on.

Answer 11

But thanks @Kainui =)

Answer 12

@Abdulhameed how?

Answer 13

do u know the derivative of arctan?

Answer 14

its du/dx / 1- u^2

Answer 15

its +

Answer 16

nope it's du/dx/1+u^2

Answer 17

yeah it is

Answer 18

u have y=(y) arctane^-2x

Answer 19

do u know what is the derivative of g(x)*f(x)?

Answer 20

nope..? i dont know?

Answer 21

derivative of
\[ f(x)*g(x)=f(x)*g \prime (x)+g(x)*f \prime(x)\]

Answer 22

the product rule. Tsk shame on me.

Answer 23

no it's ok :D everyone make mistakes

Answer 24

this equation is also an application of natural logarithms.. well, the e..

Answer 25

\[dy/dx=y \times \frac{ -2e ^{-2x} }{ 1+e ^{-4x} }+arctane ^{-2x}\times dy/dx\]

Answer 26

whats after this? =)

Answer 27

move the terms with dy/dx in one side
\[dy/dx-arctane ^{-2x} \times dy/dx=y \times \frac{ -2e ^{-2x} }{ 1+e ^{-4x} }\]

Answer 28

then take dy/dx as a factor

Answer 29

im so sorry.. can you show it to me? =)

Answer 30

\[dy/dx(1-arctane ^{-2x})=y \times \frac{ -2e ^{-2x} }{ 1+e ^{-4x}}\]

Answer 31

I've let this go on too long. This is really wrong, @silverxx you have the wrong equation written down and need to fix it since this person believes that y=0 will cause us not to divide both sides by y.

See, if y doesn't equal zero, we can divide like I described.

If y=0 then the stuff multiplied by y is completely independent of y itself.

Answer 32

So you can see in either case, arctan(e^(-2x)) is independent of y.

Answer 33

@Kainui what do u mean ?

Answer 34

I mean exactly what I said. Read it.

Answer 35

Oh no.. in the book, the equation is y= arctan e ^-2x.. and in our quiz, it is y= y arctan e^-2x

Answer 36

\[y=y*\tan^{-1}(e^{-2x})\] So if y =/= 0 then \[1=\tan^{-1}(e^{-2x})\] and we can solve for x which will be some constant.
\[y=y*C\]
the only thing that can make this true is for y=1 which is a constant (y can't be 0 since we just said above it wasn't)

if y=0 then
\[0=0*\tan^{-1}(e^{-2x})\]
In both possible cases y is either 1 or 0, which is completely independent of x.

Answer 37

im so so sorry @Abdulhameed i was wrong.. i guess its y = arctan e ^-2x

Answer 38

@Kainui you were actually right.. its y= arctan e ^-2x

Answer 39

how do we solve for this..

Answer 40

Alright, now you guys can proceed then. It's still fairly similar.

Answer 41

@silverxx oh if it's y= arctan e ^-2x then
dy/dx=-2e^-2x/1+e^-4x

Answer 42

thats it? =) even if e is natural logarithm?

Answer 43

if it's y= arctan e ^-2x then
dy/dx=-2e^-2x/1+e^-4x

Answer 44

Thank you so much @Abdulhameed and @Kainui =)))))))

Answer 45

u r welcome :D , anytime