Implicit differentiation help I want to take the second derivative of: siny + x = y. I found the first derivative to be (-1)/(cosy - 1) How do I find the second derivative?
how can you get that first derivative? show your work, please
Alright hold on
\[\frac{ dy }{ dx } [siny + x = y]\] \[cosy[\frac{dy}{dx}(1)]+\frac{d}{dx}(x)=\frac{dy}{dx}(1)\] \[cosy\frac{dy}{dx}-\frac{dy}{dx}(1)=-1\] \[\frac{dy}{dx}(cosy -1) = -1\] \[\frac{dy}{dx} = \frac{-1}{cosy-1}\]
Implicit differentiation be killing
Well can you tell me how? It seems like a different process
what did you get? i did it too not the right answer
we'll seee
sorry man that's wrong ;/
ok, so, I am helpless here. Sorry for that. let me take it off to not bother you. hehehe.
haha alright thanks for trying i'll see if i can figure it out
\[ \frac{dy}{dx} = \frac{-1}{\cos y-1} \\ \frac{dy}{dx} = \frac{1}{1- \cos y}\\ \frac{dy}{dx} =(1- \cos y)^{-1} \] now take the derivative of both sides \[ \frac{d^2 y}{dx^2} = -1(1-\cos y)^{-2} (- (-\sin y)) \frac{dy}{dx}\] replace dy/dx with \(\frac{1}{1- \cos y}\) and simplify
@phi he said that it's wrong, hehehe... I feel good when both us get the same answer
were you given the correct answer?
no, he stated that mine is wrong at few box above,
I was actually addressing @OrthodoxMan ^_^ How do you know that their answer is wrong?
I think you should get (1-cos y)^3 in the denominator
or (cos y -1)^3 depending on where you put the minus sign
Actually I gave the wrong first derivative. But phi got it right. I figured it out after so thanks for your help. Loser66 got it wrong anyways because the quotient rule isnt right And terrence, i have the answer in my textbook/
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