If y^2+_e(^2x)=y, x(-2,2), then:

Question

If $y^2+\log_e(\cos^2x)=y$, $x\in\left(-\dfrac\pi2,\dfrac\pi2\right)$, then:

Accepted Answer

At $x=0$: $y^2=y\Rightarrow y=0$ or $1$. Differentiating $y^2+2\ln\cos x=y$: $2yy'-2	an x=y'$. At $x=0$: $y'(2y-1)=0\Rightarrow y'(0)=0$.

Differentiating again: $2(y')^2+2yy''-2\sec^2x=y''$. At $x=0$, $y'=0$: $2yy''-2=y''\Rightarrow y''(2y-1)=2$. For $y=0$: $y''=-2$; for $y=1$: $y''=2$. Either way $|y''(0)|=2$.

Solutions · 1