burkinaval1b

2022-01-21

Second derivative using implicit differentiation with respect to x of $x=\mathrm{sin}y+\mathrm{cos}y$
$1=\mathrm{cos}y\frac{dy}{dx}-\mathrm{sin}y\frac{dy}{dx}$
$1=\frac{dy}{dx}\left(\mathrm{cos}y-\mathrm{sin}y\right)$
$\frac{dy}{dx}=\frac{1}{\mathrm{cos}y-\mathrm{sin}y}$

Jeffery Autrey

Expert

$\frac{dy}{dx}=\frac{1}{\mathrm{cos}y-\mathrm{sin}y}$.
Take the derivative of both sides using one of the derivative rules:
$\frac{{d}^{2}y}{{dx}^{2}}=\frac{{\left(1\right)}^{\prime }\left(\mathrm{cos}y-\mathrm{sin}y\right)-\left(1\right){\left(\mathrm{cos}y-\mathrm{sin}y\right)}^{\prime }}{{\left(\mathrm{cos}y-\mathrm{sin}y\right)}^{2}}=?$,
or, ${\left({x}^{-1}\right)}^{\prime }=-{x}^{-2}⇒\frac{{d}^{2}y}{{dx}^{2}}=\frac{-1}{{\left(\mathrm{cos}y-\mathrm{sin}y\right)}^{2}}\cdot {\left(\mathrm{cos}y-\mathrm{sin}y\right)}^{\prime }=?$
Above are the beginnings to (i) the quotient rule and (ii) the power rule and chain rules.

Marcus Herman

Expert

Just differentiate both sides of $\frac{dy}{dx}={\left(\mathrm{cos}y-\mathrm{sin}y\right)}^{-1}$ with respect to x. This leads to:
$\frac{{d}^{2}y}{{dx}^{2}}=\frac{d}{dx}\frac{dy}{dx}$
$=\frac{d}{dx}{\left(\mathrm{cos}y-\mathrm{sin}y\right)}^{-1}=-{\left(\mathrm{cos}y-\mathrm{sin}y\right)}^{-2}\cdot \left[\left(-\mathrm{sin}y\right)\frac{dy}{dx}-\left(\mathrm{cos}y\right)\frac{dy}{dx}\right]$.

RizerMix

Expert

$x=\mathrm{sin}y+\mathrm{cos}y$ first derivative by implicit diffferentiation;$1=\mathrm{cos}y\left({y}^{\prime }\right)-\mathrm{sin}y\left({y}^{\prime }\right)1={y}^{\prime }\left(\mathrm{cos}y-\mathrm{sin}y\right){y}^{\prime }=1/\left(\mathrm{cos}y-\mathrm{sin}y\right)$ second derivetive ${y}^{″}=-\left(-\mathrm{sin}y\left({y}^{\prime }\right)-\mathrm{cos}y\left({y}^{\prime }\right)\right)/\left(\mathrm{cos}y-\mathrm{sin}y{\right)}^{2}{y}^{″}=\left({y}^{\prime }\mathrm{sin}y+{y}^{\prime }\mathrm{cos}y\right)/\left(\mathrm{cos}y-\mathrm{sin}y{\right)}^{2}$ subsitute the value of y' we hwve;${y}^{″}=\left(1/\left(\mathrm{cos}y-\mathrm{sin}y\right)\left(\mathrm{sin}y+\mathrm{cos}y\right)\right)/\left(\mathrm{cos}y-\mathrm{sin}y{\right)}^{2}$${y}^{″}=\left(\mathrm{sin}y+\mathrm{cos}y\right)/\left(\mathrm{cos}y-\mathrm{sin}y{\right)}^{2}$ simplifying this we have; ${y}^{″}=-\mathrm{sin}y/2\mathrm{sin}y\mathrm{cos}y-\mathrm{cos}y/2\mathrm{sin}y\mathrm{cos}y$${y}^{″}=-\mathrm{sec}y/2-\mathrm{csc}y/2$