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#