Question

What is the equation of the tangent line of `f(x)=ln(x^2+4x+e)` at `x=0`?

Answer 1

`y=(4x)/e+1`

Explanation:

First, find the point that the tangent line will intercept:

`f(0)=ln(0^2+4(0)+e)=ln(e)=1`

Hence the point of tangency is `(0,1)`.

All we need to know now is the slope of the tangent line at `x=0`. To do this, find the value of the derivative at `x=0`.

To find the derivative of `f(x)`, we will use the chain rule.

Specifically, since

`d/dxln(x)=1/x`

We know that

`d/dxln(g(x))=1/g(x)*g'(x)=(g'(x))/g(x)`

Applying this to `f(x)=ln(x^2+4x+e)`, we see that

`f'(x)=(d/dx(x^2+4x+e))/(x^2+4x+e)=(2x+4)/(x^2+4x+e)`

The slope of the tangent line is

`f'(0)=(2(0)+4)/(0^2+4(0)+e)=4/e`

The equation of a line that passes through the point `(0,1)` with a slope of `"4/"e` is

`y=(4x)/e+1`

We can check by graphing the original function and its tangent line:

graph{(ln(x^2+4x+e)-y)(y-(4x)/e-1)=0 [-.8, 1, -2.21, 4.03]}