How do we find an equation of the tangent line to the graph of:

Question

Given the point $P(0,1)$, we are asked to find an equation of the tangent line for the following implicit function: $$y^2+ye^{x}=e^{2x}+1.$$

Answer 1

Hint: First use implicit differentiation

$$2yy'+y'e^x+ye^x=2e^{2x}.$$

Now, plug in your point $x=0$ and $y=1$ and determine $y'(0)$. If you have found the slope than you can use

$$y = y'(0)(x-x_0)+y_0$$

in which $x_0=0$ and $y_0=1$ gives you the equation of the tangent line.

Answer 2

The gradient of a function is normal to its level curves, so you can use the point-normal form of equation of a line to derive an equation of the tangent at a point $(x_0,y_0)$: $$\nabla f(x_0,y_0)\cdot(x-x_0,y-y_0)=0.\tag{*}$$

Rearrange the implicit equation to get $f:(x,y)\mapsto y^2+ye^x-e^{2x}-1$ and compute its gradient: $$\nabla f = \left(ye^x-2e^{2x},e^x+2y\right).$$ Plugging this and the given point into (*) gives $(-1,3)\cdot(x-0,y-1)=0$, or $x-3y+3=0$.