How can i calculate this tangent line, when derivate is indeterminated?

Question

i'v got this equation: $$2e^{xy}-1=x^2+y^2$$ I need the tangent line at (0,1) But when i use implicit derivate procedure i found this equation: $$\frac{x-e^{xy}}{e^{xy}-y}$$

which of course at (0,1) is $\frac{1}{0}$

what can I do?

Answer 1

One has that \begin{align*} (2e^{xy} - 1)' = (x^{2} + y^{2})' & \Longleftrightarrow 2e^{xy}(y + xy') = 2x + 2yy'\\\\ & \Longleftrightarrow e^{xy}(y + xy') = x + yy' \end{align*}

If we plug-in the values $x = 0$ and $y = 1$, one gets the relation \begin{align*} e^{0}(1 + 0 \times y') = 0 + 1\times y' \Longleftrightarrow y'(0) = 1 \end{align*}

Can you take it from here?

Answer 2

Doing implicit derivation and solving for $y'$ gives $$ 2e^{xy}-1=x^2+y^2 \\ 2e^{xy}y + 2e^{xy}xy' = 2x + 2yy' \\ 2e^{xy}xy' - 2yy' = 2x - 2e^{xy}y \\ 2(e^{xy}x-y)y' = 2(x - e^{xy}y) \\ y' = \frac{x-e^{xy}y}{e^{xy}x-y} \\ y'(x=0,y=1) = \frac{0-e^{0\cdot1}1}{e^{0\cdot1}0-1} = \frac{-1}{-1} = 1. $$

Answer 3

You appear to have made an algebraic error in solving for $y'$. On differentiating both sides with respect to $x$ we have $$2e^{xy}(y+xy')=2x+2yy'.$$ Solving this for $y'$ produces $$y'={x-ye^{xy}\over xe^{xy}-y}.$$ Compared to this expression, yours is missing a factor of $x$ in the denominator.

There’s a way to find this tangent line without explicitly computing $y'$ if you’re familiar with the gradient of a function. Define $F(x,y)=2e^{xy}-1-(x^2+y^2)$ and consider the curve $F(x,y)=0$. The gradient $\nabla F$ is normal to this curve, so we can write the equation of the tangent to it at $(0,1)$ in the point-normal form $\nabla f(0,1)\cdot(x,y-1)=0$.