1. ## Implicit differentiation

Can someone spot my mistake, if there is one? Evaluate y' at the point (-1,2) if xe^y - 6y = 2x - 10 - e^2. Here's what I did: Taking the derivative of both sides, I got e^y + xe^y(y') - 6yy' = 2. Solving for y': y' = (2-e^y)/(xe^y - 6y). Putting in the values x=-1 and y = 2, I got: y' at that point = (2-e^2)/(-e^2-12). The computer homework system I'm using says this is incorrect. Did I make a mistake somewhere? Thanks for speedy help.

2. $\displaystyle \text{Evaluate } y' \text{ at the point } (-1,2) \text{ if } xe^y - 6y = 2x - 10 - e^2$.

[math] xe^y - 6y = 2x - 10 - e^2 [/tex]
Isolate constant terms:
$\displaystyle -2x + xe^y - 6y = -10 - e^2$
Differentiate both sides:
$\displaystyle -2+[(1*e^y)+(x*e^y)y'] -6y' = 0$

$\displaystyle -2 + e^y + (xe^y)y' -6y' = 0$
$\displaystyle (xe^y)y' -6y' = 2 - e^y$
$\displaystyle (xe^y - 6) y' = 2 - e^y$

y' = $\displaystyle \frac{2-e^y}{x e^y - 6} = -1 \frac{e^y-2}{x e^y - 6}$

You made a differentiation error on the 6y term. If f(y) = 6y,
then f'(y) = 6*(y'). If this were not implicit differentiation: if f(y)=6y,
f'(y) =6.

3. Thanks - I think I was needlessly employing the chain rule there.
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