Loading...

find the implicit d… - QuestionCove
OpenStudy (anonymous):

find the implicit derivative of 2xy+y^2=x+y

2 years ago
OpenStudy (anonymous):

\[2(y+xy')+2yy'=1+y'\] us a start

2 years ago
OpenStudy (anonymous):

why is y' added on to it all?

2 years ago
OpenStudy (anonymous):

it is not added for sure are you asking why it is there?

2 years ago
OpenStudy (anonymous):

yes

2 years ago
OpenStudy (anonymous):

ok the idea is that in this curve, at least locally, \(y\) is some function of \(x\) i.e. \(y=f(x)\) even though you don't solve for it

2 years ago
OpenStudy (anonymous):

so \[ 2xy+y^2=x+y\] is the same as \[ 2xf(x)+f(x)^2=x+f(x)\]

2 years ago
OpenStudy (anonymous):

now when you take the derivative, you need the chain rule and also the product rule for example, the derivative of \[xf(x)\] is \[f(x)+xf'(x)\] by the product rule

2 years ago
OpenStudy (anonymous):

it is easier to write \[y+xy'\] same thing

2 years ago
OpenStudy (anonymous):

the derivative of \[f^2(x)\] is \[2f(x)f'(x)\] by the chain rule here we write the derivative of \(y^2\) is \(2yy'\) again the same thing

2 years ago
OpenStudy (anonymous):

if that is not clear imagine that \(y=\sin(x)\) and think of what you would have to do to take the derivative of \[2x\sin(x)+\sin^2(x)\]

2 years ago
OpenStudy (anonymous):

you would need the product and chain rule to do them both \[\LaTeX\] practice is all

2 years ago
OpenStudy (anonymous):

I'm still confused on how you would solve this

2 years ago
OpenStudy (anonymous):

d/dx (2xy + y^2 = x + y) use product rule for 2xy part: (derivative of the 1st * 2nd as is) + (derivative of the 2nd * 1st as is) so, [ d/dx (2x) * y ] + [ d/dx y * (2x) ] = [2*y] + [ (dy/dx) * 2x] That's for the 2xy part. derivative of y^2 is 2y dy/dx because you move the exponent to the front of the expression, reduce the exponent by subtracting 1, and stick dy/dx at the end. derivative of x is just 1 because the exponent is understood to be 1 so put that in the front of the expression, then subtract the exponent of 1 by 1 to get 0, anything to the 0 power is just 1. dx/dx is just 1/1=1 so you don't need to write it derivative of y is dy/dx...same explanation as d/dx of x but you have to keep dy/dx part. So, the implicit differentiation for the whole thing is: 2y + 2x(dy/dx) + 2y(dy/dx) = 1 + dy/dx Then, move all dy/dx to one side and the terms w/o that to the other side: 2x(dy/dx) + 2y (dy/dx) -(dy/dx) = 1 - 2y Factor out dy/dx: (dy/dx)[2x + 2y - 1] = 1-2y Keep just dy/dx part and move the rest over to the other side by division: dy/dx = (1-2y)/(2x+2y-1) This is your final answer!

2 years ago
Similar Questions: