Question

If y=tan u, u=v-1/v, and v=ln x, what's the value of dy/dx at x=e? (Answer: 2/e) (y=tan(ln x-1/ln x) but how do I simplify this so I can take the derivative?)

Answer 1

\[\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dv}\dfrac{dv}{dx}\]
\[\dfrac{dv}{dx}=\dfrac{1}{x}\]
\[\dfrac{du}{dv}=v+\dfrac{1}{v^2}=lnx+\dfrac{1}{ln^2x}\]
\[\dfrac{dy}{du}=sec^2u = sec^(v-\dfrac{1}{v}=sec^2(lnx-\dfrac{1}{lnx})\]
time them to get the answer -------------------------------------------------
\[sec^2(lnx-\dfrac{1}{lnx})(lnx+\dfrac{1}{ln^2x})(\dfrac{1}{x})\] and replace x =e
sec^2 (0)(2) = 2 sec^2(0) = 2