The tangent equation of the curve y = ln (x + 1) at (0,0) is

f(x)=ln(x＋1)
Then:
f'(x)=1/(x＋1)
The tangent slope is k = f '(0) = 1
The tangent point is (0,0)
Then the tangent equation is:
x－y=0

The tangent equation of curve y = x (3 ln x + 1) at point (1.1) is

Solution
Y = (3lnx + 1) is that it?
Let's first derive y
y‘=3/x
The tangent slope is k = 3 / 1 = 3
The tangent equation is Y-1 = 3 (x-1)
Namely
y=3x-2

Let f (x) = g (x) + x2 and the tangent equation of the curve y = g (x) at x = 1 be y = 2x + 1, then f (1) + F '(1) = ()
A. 6B. 7C. 8D. 9

The tangent equation of ∵ curve y = g (x) at point (1, G (1)) is y = 2x + 1, ∵ g ′ (1) = 2, G (1) = 3, ∵ function f (x) = g (x) + X2, ∵ f ′ (x) = g ′ (x) + 2x ∵ f ′ (1) = g ′ (1) + 2 ∵ f ′ (1) = 2 + 2 = 4, f (1) = g (1) + 1 = 4, ? f (1) + F ′ (1) = 8

Function f (x) = x & # 178; + 2, find the tangent equation at x = 3

Seeking derivative
f'(x)=2x
Then the slope k = f '(3) = 6
f(3)=9+2=11
The tangent point is (3,11)
So the tangent is 6x-y-7 = 0

The tangent equation of the curve y = ln (x + 1) at (0,0) is