laminarskq2p

2023-02-21

How to find ((d^2)y)/(dx^2)?

Camuccinirk84

Defining
$f\left(x,y\left(x\right)\right)=2xy\left(x\right)+2y{\left(x\right)}^{2}-13=0$ then
$\frac{df}{dx}=2y+2xy\prime +4yy\prime =0$ and
$\frac{d}{dx}\frac{df}{dx}=4y\prime +2xy\prime \prime +4y{\prime }^{2}+4yy\prime \prime =0$
after substitution of $y\prime$ we have
$y\prime \prime =\frac{2xy+2{y}^{2}}{{\left(x+2y\right)}^{3}}=\frac{13}{{\left(x+2y\right)}^{3}}$
but $y=\frac{1}{2}\left(-x±\sqrt{26+{x}^{2}}\right)$ so finally
$y\prime \prime =±\frac{13}{{\left(26+{x}^{2}\right)}^{\frac{3}{2}}}$

Ruby Rollins

$2xy+2{y}^{2}=13$
Differentiating wrt $x$ and applying the product rule gives us:
$2\left\{\left(x\right)\left(\frac{dy}{dx}\right)+\left(1\right)\left(y\right)\right\}+4y\frac{dy}{dx}=0$
$x\frac{dy}{dx}+y+2y\frac{dy}{dx}=0⇒\frac{dy}{dx}=-\frac{y}{x+2y}$
Differentiating again wrt $x$ and applying the product rule (twice) gives us:
$\therefore \left\{\left(x\right)\left(\frac{{d}^{2}y}{{dx}^{2}}\right)+\left(1\right)\left(\frac{dy}{dx}\right)\right\}+\frac{dy}{dx}+2\left\{\left(y\right)\left(\frac{{d}^{2}y}{{dx}^{2}}\right)+\left(2\frac{dy}{dx}\right)\left(\frac{dy}{dx}\right)\right\}=0$
$\therefore x\frac{{d}^{2}y}{{dx}^{2}}+\frac{dy}{dx}+\frac{dy}{dx}+2y\frac{{d}^{2}y}{{dx}^{2}}+2{\left(\frac{dy}{dx}\right)}^{2}=0$
$\therefore x\frac{{d}^{2}y}{{dx}^{2}}+2\frac{dy}{dx}+2y\frac{{d}^{2}y}{{dx}^{2}}+2{\left(\frac{dy}{dx}\right)}^{2}=0$
$\therefore \left(x+2y\right)\frac{{d}^{2}y}{{dx}^{2}}+2\left(-\frac{y}{x+2y}\right)+2{\left(-\frac{y}{x+2y}\right)}^{2}=0$
$\therefore \left(x+2y\right)\frac{{d}^{2}y}{{dx}^{2}}=\frac{2y}{x+2y}-\frac{2{y}^{2}}{{\left(x+2y\right)}^{2}}$
$\therefore \frac{{d}^{2}y}{{dx}^{2}}=\frac{2y}{{\left(x+2y\right)}^{2}}-\frac{2{y}^{2}}{{\left(x+2y\right)}^{3}}$
Overlaying a common denominator results in:
$\frac{{d}^{2}y}{{dx}^{2}}=\frac{\left(2y\right)\left(x+2y\right)-\left(2{y}^{2}\right)}{{\left(x+2y\right)}^{3}}$
$\therefore \frac{{d}^{2}y}{{dx}^{2}}=\frac{\left(2xy+4{y}^{2}-2{y}^{2}\right)}{{\left(x+2y\right)}^{3}}$
$\therefore \frac{{d}^{2}y}{{dx}^{2}}=\frac{\left(2xy+2{y}^{2}\right)}{{\left(x+2y\right)}^{3}}$
$\therefore \frac{{d}^{2}y}{{dx}^{2}}=\frac{13}{{\left(x+2y\right)}^{3}}$ (as $2xy+2{y}^{2}=13$)

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