Question

If $e^y(x+1)=1$, show that $\frac{dy}{dx}=-e^y$.

Answer 1

$e^y(x+1)=1e^y=\frac{1}{1+x}⟶\left(1\right)\ln e^y=\ln \left(\frac{1}{1+x}\right)y=\ln {(x+1)}^{-1}=-\ln (x+1)\frac{dy}{dx}=-\frac{d\ln (x+1)}{dx}=-\frac{1}{x+1}$

From $\left(1\right)$ we get $\frac{dy}{dx}=-e^y$

$x^2-6\sqrt{3}+28-{\sin }^2\left(\frac{\pi x}{6\sqrt{3}}\right)=0x^2-23\sqrt{3}x+27+1-{\sin }^2\left(\frac{\pi x}{6\sqrt{3}}\right)=0{(x-3\sqrt{3})}^2+{\cos }^2\left(\frac{\pi x}{6\sqrt{3}}\right)=0$

As both are squares so both $=0$

$x=3\sqrt{3}$ & $\cos \left(\frac{\pi x}{6\sqrt{3}}\right)=0$

At $x=3\sqrt{3}\cos \frac{\pi x}{6\sqrt{3}}=0$

&$x=3\sqrt{3}\cos \frac{\pi }{2}=0$

There are $3$ solution. Answer A