Please help the equation of the curve is x lny = 2x + 1 the question asks to show that dy/dx = -y/x^2 i only got as far as : lny + x/y dy/dx = 2 now what ? thank you :D
Try divide both sides by x then take the derivative.
\[x \ln y = 2x + 1 \Rightarrow \ln y = \dfrac{2x + 1}{x}\] \[\dfrac{\text{d}}{\text{d}x}\ln y = \dfrac{\text{d}}{\text{d}x}\left(\dfrac{2x + 1}{x} \right) \] \[\dfrac{1}{y} \cdot \dfrac{\text{d}y}{\text{d}x} = \dfrac{\text{d}}{\text{d}x}\left(\dfrac{2x + 1}{x} \right) \]
if i divide both sides by x i get 1/y dy/dx = 2/x - 2x -1
ok so how do i get that to equal
try looking at the steps @geerky42 have done.. your question is this. x lny = 2x + 1 just divide both sides by x
Maybe it won't work. All you have to do is to make equation clean and simple. From that, you probably will find a way to prove that dy/dx = -y/x².
Make it simple, then take the derivative of it then simplify it.
hmm @geerky42 are u saying your method won't work the one u have mentioned or what?
I don't know. Try it and figure it out.
it would work i have tried it..
If it doesn't work, find another way to approach to solution.
nubeer you got -y/x^2 ?
yup.. @geerky42 approach of solving the question is right.. just keep simplifying same way he has done.
Ok. Thanks. Do you know what to do now? @1234abdc
no, I'm really confused
Just divide both sides by x so you won't have ugly and scary-look equation when you derivative it.
Do you know how to take the derivative, right?
yes i do. my problem is simplifying
Well, can you do it from here?
\[\frac{ 1 }{ y } \frac{ dy }{ dx } = \frac{ 2 }{ x } - 2x - 1\]
now what ?
None, it's not correct. Do you know quotient rule, right?
yes.. why is it wrong ?
I don't know how you got \( \dfrac{2}{x}-2x-1 \), but I got \( \dfrac{2x - (2x+1)}{x^2} \) which can be simplifed to \( -\dfrac{1}{x^2} \).
oh ! i just got it i made a really stupid mistake thank you for your help :D
You're welcome. Glad I helped.
agreed
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