Faith Welch

2022-07-17

So my problem is that I have to use implicit differentiation to find the derivative of $F\left(x,y\right)={e}^{xy}-x$ when $F\left(x,y\right)=10$ and the equation of the tangent at the point $\left(1,\mathrm{log}\left(11\right)\right)$. 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 $\mathrm{ln}$ to simply into $\mathrm{ln}\left(10+x\right)=xy$ 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?

Danica Ray

Expert

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.

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

Do you have a similar question?