If and , find by implicit differentiation.
$\displaystyle \frac{1}{x} + \frac{1}{y} = 2$
$\displaystyle x^{-1} + y^{-1} = 2$
$\displaystyle \frac{d}{dx}(x^{-1} + y^{-1}) = \frac{d}{dx}(2)$
$\displaystyle -x^{-2} + \frac{d}{dy}(y^{-1})\,\frac{dy}{dx} = 0$
$\displaystyle -\frac{1}{x^2} - \frac{1}{y^2}\,\frac{dy}{dx} = 0$
$\displaystyle -\frac{1}{y^2}\,\frac{dy}{dx} = \frac{1}{x^2}$
$\displaystyle \frac{dy}{dx} = -\frac{y^2}{x^2}$.
You also know that when $\displaystyle x = 4, y = \frac{4}{7}$.
So $\displaystyle \frac{dy}{dx}|_{x = 4} = -\frac{\left(\frac{4}{7}\right)^2}{4^2}$
$\displaystyle = -\frac{\frac{16}{49}}{16}$
$\displaystyle = -\frac{1}{49}$.