Find $y'' of$
$9x^2 + y^2 = 7$
I've been working on this one for a while now, I'm not sure where I'm going wrong. I posted an image of my work, I'm hoping someone can take a look at it and give me some feedback
2 Answers
Notice, We have $$9x^2+y^2=7$$ differentiating both the sides w.r.t. $x$, we get $$\frac{d}{dx}(9x^2+y^2)=\dfrac{d}{dx}(7)$$ $$18x+2y\frac{dy}{dx}=0$$ $$\frac{dy}{dx}=\frac{-9x}{y}\tag 1$$ Differentiating (1) w.r.t. $x$, we get $$\frac{d^2y}{dx^2}=\frac{y(-9)+9x\frac{dy}{dx}}{y^2}$$ setting the value of $\frac{dy}{dx}$ from (1), we get $$\frac{d^2y}{dx^2}=\frac{y(-9)+9x\frac{(-9x)}{y}}{y^2}$$ $$\color{red}{y''=\frac{d^2y}{dx^2}=-\frac{9(9x^2+y^2)}{y^3}}$$
$$\frac{d}{dx}(9x^2+y^2)=\frac{d}{dx}(7)$$ $$18x+2yy'=0$$
$$\frac{d}{dx}(18x+2yy')=0$$ $$(18+2y'y'+2yy'')=0$$ $$y''=\frac{-18-2y'y'}{2y}=\frac{-9-y'^2}{y}$$ so that $$y'=\frac{-18x}{2y}=\frac{-9x}{y}$$
