Thread: Find the slope of the curve (derive) ln(x^2+1) at point (0,0)

This is what I have so far for the problem: derive ln(x^2+1) at (0,0)

y=ln(u) u=x^2+1

y=1/x(u)*2x

y=1/x(x^2+1)*2x

y=2(x^2+1)

0=2(x^2+1)

0=(x^2+1)

I'm not sure how to factor x^2+1. I was wondering if i'm doing this problem correctly. If so, should I just use the quadratic equation for that? I think that I may get an imaginary number.

Thanks! This forum seems legit.

-star

Originally Posted by starbless

This is what I have so far for the problem: derive ln(x^2+1) at (0,0)

y=ln(u) u=x^2+1

y=1/x(u)*2x

y=1/x(x^2+1)*2x

y=2(x^2+1)

0=2(x^2+1)

0=(x^2+1)

I'm not sure how to factor x^2+1. I was wondering if i'm doing this problem correctly. If so, should I just use the quadratic equation for that? I think that I may get an imaginary number.

Thanks! This forum seems legit.

-star