If x_e(_ex)-x^2+y^2=4 (y0), then dydx at x=e is equal to:

Question

If $x\log_e(\log_ex)-x^2+y^2=4$ ($y>0$), then $\dfrac{dy}{dx}$ at $x=e$ is equal to:

Accepted Answer

Differentiating: $\ln(\ln x)+\dfrac{1}{\ln x}-2x+2y\dfrac{dy}{dx}=0$.

At $x=e$: $y=\sqrt{4+e^2}$ (from the original equation). Substituting $\ln e=1$: $0+1-2e+2\sqrt{4+e^2}\dfrac{dy}{dx}=0\Rightarrow\dfrac{dy}{dx}=\dfrac{2e-1}{2\sqrt{4+e^2}}$.

Solutions · 1