# How do you use implicit differentiation to find the slope of the line tangent to y+ lnxy=4 at (.25, 4)?

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{16}{5}$ =[3.2] [The gradient]
$y + \ln x y = 4$, so by the theory of logs, $y + \ln x + \ln y = 4$
Differentiating both sides of the equation implicitly with respect to x,....... $\frac{\mathrm{dy}}{\mathrm{dx}} + \frac{1}{x} + \frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$, Therefore,
$\frac{\mathrm{dy}}{\mathrm{dx}} \left[1 + \frac{1}{y}\right] = - \frac{1}{x}$...... so $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{y}{x \left[y + 1\right]}$ and substituting for $x = 0.25 \mathmr{and} y = 4$ will give the above result for the gradient.