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------------------------------------------------ Find an equation of the tangent line to the curve [tex]$y=3 x^3$[/tex] at the point [tex]$(1,3)$[/tex].

A. [tex]y=6 x-18[/tex]
B. [tex]y=9 x-6[/tex]
C. [tex]y=6 x-9[/tex]
D. [tex]y=9 x-12[/tex]
E. [tex]y=3 x-9[/tex]

Answer :

Sure! Let's find the equation of the tangent line to the curve [tex]\(y = 3x^3\)[/tex] at the point [tex]\((1, 3)\)[/tex].

1. Find the derivative of [tex]\(y\)[/tex] to get the slope of the tangent line:

The curve is given by:
[tex]\[
y = 3x^3
\][/tex]

The first derivative of [tex]\(y\)[/tex] with respect to [tex]\(x\)[/tex] (which represents the slope of the tangent line) is:
[tex]\[
\frac{dy}{dx} = \frac{d}{dx}(3x^3) = 9x^2
\][/tex]

2. Evaluate the derivative at [tex]\(x = 1\)[/tex] to find the slope at this point:

Substitute [tex]\(x = 1\)[/tex] into the derivative:
[tex]\[
\left. \frac{dy}{dx} \right|_{x = 1} = 9(1)^2 = 9
\][/tex]

So, the slope of the tangent line at [tex]\(x = 1\)[/tex] is [tex]\(9\)[/tex].

3. Use the point-slope form of a line to find the equation of the tangent line:

The point-slope form of a line is given by:
[tex]\[
y - y_1 = m(x - x_1)
\][/tex]

where [tex]\((x_1, y_1)\)[/tex] is the point on the line and [tex]\(m\)[/tex] is the slope. Here, the point is [tex]\((1, 3)\)[/tex] and the slope [tex]\(m\)[/tex] is [tex]\(9\)[/tex].

Substitute these values into the point-slope form:
[tex]\[
y - 3 = 9(x - 1)
\][/tex]

4. Simplify the equation to standard form:

Expand and simplify:
[tex]\[
y - 3 = 9x - 9
\][/tex]

Adding [tex]\(3\)[/tex] to both sides:
[tex]\[
y = 9x - 6
\][/tex]

Therefore, the equation of the tangent line to the curve [tex]\(y = 3x^3\)[/tex] at the point [tex]\((1, 3)\)[/tex] is:
[tex]\[
y = 9x - 6
\][/tex]

So, the correct answer is:
[tex]\[
y = 9x - 6
\][/tex]