Question #ca7f8

2 Answers
May 20, 2015

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]}

May 20, 2015

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