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

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?

$$F(x,y) = 10$$ 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 $$\frac{dy}{dx}\approx -2.307.$$