Implicit Differentiation of ln y = 2 x y^2

[matsh3]

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.

 

[stapel]

...implicit differentiation.

ln y = 2xy^2

[explanation and worked example snipped]

Ne'mind; hand-in solution posted below.

Eliz.

 

[soroban]

Hello, matsh3!

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

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

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

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

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

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