Equation of tangent line y = ln(x^2 + 1)

Question

at the point (1, ln2)?

Answer 1

y = log(x² + 1)

Differenciate each side with respective x

dy / dx = 1 / (x² + 1) [2x + 0]

            = 2x / x² + 1

The slope of the tangent line m = 2x / x² + 1 at point ( 1 , log2)

Therefore m = 2(1) / 1 + 1 = 2 / 2 = 1

The tangent line equation formula : y - y1 = m(x - x1)

Substitute m = 1 , x 1 = 1 and y1 = log2 in the above tangent line

y - log2 = 1(x - 1)

y - log2 = x - 1

Add 1 to each side

y  + 1 - log2 = x

x - y = 1 - log2

x - y = 1 - 0.3010

x - y = 0.699

x - y = 699 / 1000

1000x - 1000y = 699.