# Math Help - Second derivative using implicit differentiation?

1. ## Second derivative using implicit differentiation?

Find d^2/dx^2 (second derivative) if y^2+2y=4x^2+2x.
The first derivative is 4x+1/y+1. Now, when I get to 4(y+1)-(4x+1)(4x+1/y+1) / (y+1)^2, I'm stuck. My book says the answer is 4(y+1)^2-(4x+1)^2 / (y+1)^3 is the answer... how do you get that answer? What must you do? And why does the denominator increase by a power?

\displaystyle \begin{align*} y^2 + 2y &= 4x^2 + 2x \\ \frac{d}{dx}\left(y^2 + 2y\right) &= \frac{d}{dx}\left(4x^2 + 2x\right) \\ \left(2y + 2\right)\frac{dy}{dx} &= 8x + 2 \\ \left(y + 1\right)\frac{dy}{dx} &= 4x + 1 \\ \frac{d}{dx}\left[\left(y + 1\right)\frac{dy}{dx}\right] &= \frac{d}{dx}\left(4x + 1\right) \\ \left(y + 1\right)\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 &= 4 \\ \left(y + 1\right)\frac{d^2y}{dx^2} + \left(\frac{4x + 1}{y + 1}\right)^2 &= 4 \\ \left(y + 1\right)\frac{d^2y}{dx^2} + \frac{(4x + 1)^2}{(y + 1)^2} &= 4 \\ \left(y + 1\right)\frac{d^2y}{dx^2} &= 4 - \frac{(4x+1)^2}{(y+1)^2} \\ \left(y + 1\right)\frac{d^2y}{dx^2} &= \frac{4(y + 1)^2 - (4x + 1)^2}{(y + 1)^2} \\ \frac{d^2y}{dx^2} &= \frac{4(y + 1)^2 - (4x + 1)^2}{(y + 1)^3} \end{align*}
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