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

• Feb 7th 2010, 01:29 PM
starbless
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
• Feb 7th 2010, 01:47 PM
skeeter
Quote:

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

$\frac{d}{dx} [\ln{u}] = \frac{1}{u} \cdot \frac{du}{dx}$

$\frac{d}{dx} [\ln(x^2+1)] = \frac{1}{x^2+1} \cdot 2x = \frac{2x}{x^2+1}$