Question

lf $y=\sqrt{\cos 2x}$, then $y\frac{d^{2}y}{d{x}^2}+2y^2=$

• A. $0$
• B. $-(\frac{dy}{dx}{)}^2$
• C. $(\frac{dy}{dx}{)}^2$
• D. $y\frac{dy}{dx}$

Answer 1

The correct option is B $-(\frac{dy}{dx}{)}^2$

$y^{{}^{\prime }}=\frac{-sin2x}{\sqrt{cos2x}}$
$y{y}^{{}^{\prime }}=-sin2x$
$(y^{{}^{\prime }}{)}^2+y{y}^{{}^{\prime \prime }}=-2cos2x$
$(y^{{}^{\prime }}{)}^2+y{y}^{{}^{\prime \prime }}=-2y^2$
$y{y}^{{}^{\prime \prime }}+2y^2=-(y^{{}^{\prime }}{)}^2$