Second derivative using implicit differentiation with respect to x of x=siny+cosy 1=cosydydx−sinydydx 1=dydx(cosy−siny) dydx=1cosy−siny

burkinaval1b

Answered

2022-01-21

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

Answer & Explanation

Jeffery Autrey

Expert

2022-01-21Added 35 answers

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

Marcus Herman

Expert

2022-01-22Added 41 answers

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

RizerMix

Expert

2022-01-27Added 437 answers

$x=\mathrm{sin}y+\mathrm{cos}y$ first derivative by implicit diffferentiation;$1=\mathrm{cos}y({y}^{\prime})-\mathrm{sin}y({y}^{\prime})1={y}^{\prime}(\mathrm{cos}y-\mathrm{sin}y){y}^{\prime}=1/(\mathrm{cos}y-\mathrm{sin}y)$ second derivetive ${y}^{\u2033}=-(-\mathrm{sin}y({y}^{\prime})-\mathrm{cos}y({y}^{\prime}))/(\mathrm{cos}y-\mathrm{sin}y{)}^{2}{y}^{\u2033}=({y}^{\prime}\mathrm{sin}y+{y}^{\prime}\mathrm{cos}y)/(\mathrm{cos}y-\mathrm{sin}y{)}^{2}$ subsitute the value of y' we hwve;${y}^{\u2033}=(1/(\mathrm{cos}y-\mathrm{sin}y)(\mathrm{sin}y+\mathrm{cos}y))/(\mathrm{cos}y-\mathrm{sin}y{)}^{2}$${y}^{\u2033}=(\mathrm{sin}y+\mathrm{cos}y)/(\mathrm{cos}y-\mathrm{sin}y{)}^{2}$ simplifying this we have;${y}^{\u2033}=(\mathrm{sin}y+\mathrm{cos}y)/(-\mathrm{sin}2y)\text{}\text{or}\text{}{y}^{\u2033}=-(\mathrm{sin}y+\mathrm{cos}y)/2\mathrm{sin}y\mathrm{cos}y$${y}^{\u2033}=-\mathrm{sin}y/2\mathrm{sin}y\mathrm{cos}y-\mathrm{cos}y/2\mathrm{sin}y\mathrm{cos}y$${y}^{\u2033}=-\mathrm{sec}y/2-\mathrm{csc}y/2$