Implicit Differentiation of ln y = 2 x y^2

matsh3

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I am having a lot of trouble with certain aspects of implicit differentiation. Currently I am reviewing and am struggling on this specific problem. Can anyone help me?? I am supposed to find y' (y prime)

ln y = 2xy^2
USe Implicit Differentiation to find y' (aka y prime)

Is the product rule used or the chain rule and if so how?? I am using this as an example so please relate answer to this problem.
 
matsh3 said:
...implicit differentiation.

ln y = 2xy^2
[explanation and worked example snipped]

Ne'mind; hand-in solution posted below.

Eliz.
 
Hello, matsh3!

Differentiate: .lny  =  2xy2\displaystyle \ln y \;=\;2xy^2

We have:   1y(dydx)  =  4xy(dydx)+2y2\displaystyle \text{We have: }\;\frac{1}{y}(\tfrac{dy}{dx}) \;=\;4xy(\tfrac{dy}{dx}) + 2y^2

Multiply by y ⁣:    dydx  =  4xy2(dydx)+2y3\displaystyle \text{Multiply by }y\!:\;\;\tfrac{dy}{dx} \;=\;4xy^2(\tfrac{dy}{dx}) + 2y^3

. . . dydx4xy2(dydx)  =  2y3\displaystyle \tfrac{dy}{dx} - 4xy^2(\tfrac{dy}{dx}) \;=\;2y^3

\(\displaystyle \text{Factor: }\;\tfrac{dy}{dx}\left(1 - 4xy^2) \;=\;2y^3\)

Therefore:   dydx  =  2y314xy2\displaystyle \text{Therefore: }\;\frac{dy}{dx} \;=\;\frac{2y^3}{1-4xy^2}

 
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