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

Question

William Abdalla · Accepted Answer

#dy/dx=-16/5# =#[3.2#] [The gradient]

#y+lnxy=4#, so by the theory of logs, #y+ lnx+lny=4# Differentiating both sides of the equation implicitly with respect to x,....... #dy/dx+1/x+1/ydy/dx=0#, Therefore, #dy/dx[1+1/y]=-1/x#...... so #dy/dx=-y/[x[y+1]]# and substituting for #x=0.25 and y=4# will give the above result for the gradient.

Ezekiel Alexander · Answer

undefined To find the slope of the tangent line to $ y + \ln(xy) = 4 $ at the point $ (0.25, 4) $ using implicit differentiation, follow these steps:

1. Differentiate both sides of the equation with respect to $ x $.
2. Treat $ y $ as a function of $ x $ and apply the chain rule when differentiating terms involving $ y $.
3. Solve for $ \frac{{dy}}{{dx}} $.
4. Plug in the coordinates of the given point $ (0.25, 4) $ into the derivative to find the slope of the tangent line.

Starting with the equation $ y + \ln(xy) = 4 $, differentiate both sides with respect to $ x $:

$$ \frac{{d}}{{dx}} \left( y + \ln(xy) \right) = \frac{{d}}{{dx}} (4) $$

$$ \frac{{dy}}{{dx}} + \frac{1}{{xy}}(xy' + y) = 0 $$

$$ \frac{{dy}}{{dx}} + \frac{{y'}}{{x}} + \frac{{y}}{{x}} = 0 $$

Now, solve for $ \frac{{dy}}{{dx}} $:

$$ \frac{{dy}}{{dx}} = - \frac{{y}}{{x}} - \frac{{y'}}{{x}} $$

We are given the point $ (0.25, 4) $. Plug these values into the derivative:

$$ \frac{{dy}}{{dx}} = - \frac{{4}}{{0.25}} - \frac{{y'}}{{0.25}} $$

$$ \frac{{dy}}{{dx}} = -16 - 4y' $$

We can solve for $ y' $ by plugging in the given point:

$$ 4 = -16 - 4y' $$

$$ 4y' = -16 - 4 $$

$$ 4y' = -20 $$

$$ y' = -5 $$

Therefore, the slope of the tangent line to $ y + \ln(xy) = 4 $ at the point $ (0.25, 4) $ is $ -5 $.