Find the tangent line

Question

How would I go about solving for the tangent line of f(x)=x+y^2=3^x+2x at x=0??

Answer 1

Use implicit differentiation.

First, find the y value when x = 0. So, plugging in x = 0 we get:

y(0) + y² = e⁰ + 2(0)

y² = 1

y = either +1 or -1

So, the points (0, 1) and (0, -1) will be points on this graph and both will have a tangent line. Notice that these are also the y-intercepts.

We use implicit differentiation to find the slope of the tangent lines at these points:

The derivative with respect to x is:

y'x + yx' + 2y•y' = (e^x)' + (2x)'

(dy/dx)•x + y + 2y•dy/dx = e^x + 2

(dy/dx)•x + 2y•dy/dx = e^x + 2 - y

(dy/dx)(x + 2y) = e^x + 2 - y

dy/dx = (e^x + 2 - y)/(x + 2y)

Plugging in (0, 1), we get:

dy/dx = (e⁰ + 2 - 1)/(0 + 2•1)

dy/dx = (2)/(2) = 1

So that tangent line will be y = x + 1

Plugging in (0, -1), we get:

dy/dx = (e⁰ + 2 - -1)/(0 + 2•(-1))

dy/dx = (4)/(-2) = -2

So that tangent line will be y = -2x - 1