Using implicit differentiation to find equation of a tangent at a point

Question

So my problem is that I have to use implicit differentiation to find the derivative of $F(x, y) = e^{xy} - x$ when $F(x, y) = 10$ and the equation of the tangent at the point $(1, \log(11))$. So I tried to solve this using two ways:

The first way I used was rearranging the equation to $10 + x = e^{xy}$ and then using $\ln$ to simply into $\ln(10 + x) = xy$ before using implicit differentiation. The result is $$\frac{dy}{dx} = \frac{1}{10x+x^2} - \frac{y}{x}$$ and after substituting the point, I get approximately $-0.950$ as the slope.

The second way I used was just differentiating it straight away rather than rearranging and I get $$\frac{dy}{dx} =\frac{\frac{1}{e^{xy}} - y}{x}.$$ But when I sub the point in I get approximately $-0.688$ as the slope. So I'm not sure if I've done something wrong when getting the derivatives or am I not allowed to rearrange the equation?

Answer 1

Careful, I edited what I thought was a typo, but it turned out to be the source of your error, so I reversed the edit.

$F(x,y) = 10$ when $(x,y) = (1,\ln(11))$, not when $(x,y) = (1,\log(11))$, where I (and your calculator) are using $\log(11)$ to mean the base-$10$ logarithm, not the base-$e$ logarithm. Both your derivatives are correct, and when you plug in $(1,\ln(11))$ they both give you $$\frac{dy}{dx}\approx -2.307.$$