What is the equation of the tangent line to the curve f(x)=3x^2+1, at the point x=1?

We'll recall the fact that the derivative of a function at
a given point is represented by the value of the slope of the tangent line at the
curve.

The equation of the tangent line, in the point x =
1 is:

y - f(1) =
f'(1)(x-1)

We'll calculate f(1), by substituting x by 1 in
the expression of the function:

f(1) = 3*1^2 +
1

f(1) = 3 + 1

f(1) =
4

To calculate f'(1), first we'll have to differentiate the
given function with respect to x:

f'(x) = (3x^2 +
1)'

f'(x) = 6x

Now, we'll
replace x by 1 in the expression of the first
derivative:

f'(1) = 6

Now,
we'll substitute f(1) and f'(1) in the expression of the equation of the tangent
line:

y - f(1) = f'(1)(x-1)

y
- 4 = 6(x - 1)

We'll remove the
brackets:

y - 4 = 6x - 6

We'll
add 6 both sides:

y = 6x - 6 +
4

y = 6x - 2

The
equation of the tangent line, to the curve f(x) = 3x^2 + 1, is: y = 6x -
2.