Question

Question #ca7f8

Answer 1

So you don't know the derivative of `ln(x)`, right?

(I assume the function is `f(x)=1/2xln(x^2)`, otherwise `-1 !in D(f)`, so `f(x)=xln(abs(x))`, but please in the future write LaTeX-ly)

Let's write the definition:

`lim_(h->0)(ln(x+h)-ln(x))/h=lim_(h->0)ln(1+h/x)/h`

You know `lim_(a->0)ln(1+a)/a=1`, so

`lim_(h->0)ln(1+h/x)/h=lim_(h->0)(h/x)/h=1/x`

So, by Leibniz

`f'(x)=ln(abs(x)) + x1/x=ln(abs(x))+1`

So the equation of the tangent line is

`y-y_0=f'(x_0)(x-x_0)`

Which means

`y=(1+ln(1))(x+1)=x+1`

graph{1/2xln(x^2) [-10, 10, -5, 5]}
graph{y=x+1 [-10, 10, -5, 5]}

Answer 2

For `f(x)=1/2xlnx^2`,

to find the slope of the tangent line at `(-1,0)`, we need `f'(-1)`.

Usually, I like to use the fact that `ln x^r = rlnx`, but that assumes that `x > 0`, which we will not have for the point `(-1,0)`, so leave it as is.

To find `f'(x)` we'll use the product rule:

`f'(x) = 1/2 lnx^2 +1/2 x [1/x^2 *2x]`

`("Note: "d/dx(lnu) = 1/u (du)/dx)`

So, `f'(x) = 1/2 lnx^2 +1` .

And at the point of interest: `f'(-1) = 1/2 ln 1 +1 =0+1=1`

I assume you can find the equation of the line through `(-1,0)` with slope `m=1`.

Additional Comments

We could have used `lnx^2 = 2 ln abs(x)` together with

`d/dx ln abs x = 1/x` to get

`d/dx(lnx^2) = d/dx(2ln abs x ) = 2d/dx(ln abs x ) = 2/x`