Question

How do you write the equation of the tangent line to the graph of `y=ln(x^2+e)` at x=0?

Answer 1

Equation of the tangent at `x=0` is `y=1`

Explanation:

`y= ln(x^2+e)`

`dy/dx = 1/(x^2+e) * 2x`

Therefore slope of the tangent to `y` at `x=0` is:
`m = 1/e * 0 = 0`

Equation of the tangent is given by: `y=mx+c` Where `m` is the slope and `c` is constant

Since `m = 0 -> y = c` In this case

From original function: `y(0) = ln(0+e) = ln(e) = 1`
Therefore the tangent touches `y` at the point `(0,1)`

Hence `c=1`

Therefore the equation of the tangent at `x=0` is `y=1`
(a horizontal straight line where `y=1` for all `x`)