1. ## tangent and normal

Hi i wonder what the wrong is in my calculation is :
deside the tangent and the normal to the curve f(x) = ln(x + sqrt(1 + x^2) in the pont (0,0).

f´(x)= 1/(sqrt(x^2 + 1))
f´(0) = 1/(sqrt(0^2 + 1)) = 1
y = kx + m, y = 0, x = 0, k = 1
m = y -kx = 1 - 1 * 0 = 0

but I must have done wrong somewere, for the right answer is tangen, y= x, and the normal, y = -x.

help someone :/

2. Originally Posted by paulaa
Hi i wonder what the wrong is in my calculation is :
deside the tangent and the normal to the curve f(x) = ln(x + sqrt(1 + x^2) in the pont (0,0).

f´(x)= 1/(sqrt(x^2 + 1))
f´(0) = 1/(sqrt(0^2 + 1)) = 1
y = kx + m, y = 0, x = 0, k = 1
m = y -kx = 1 - 1 * 0 = 0

but I must have done wrong somewere, for the right answer is tangen, y= x, and the normal, y = -x.

help someone :/
1. This is definitely the wrong forum!

2. You solved the question perfectly - honestly!

You got: k = 1 and m = 0. Therefore the equation of the tangent is:

$y = 1 \cdot x + 0~\implies~\boxed{y=x}$

3. The normal has the slope $k_n=-\frac11$ and passes through the tangent point T(0, 0).