find equation of tangent line to y = (x^2 + 1)^3

[yummymummy1713]

I need to find the equation of a tangent line to the curve y = (x^2 + 1)^3 at the point (1, 8).

I know that, for starters, I need to take the derivative. I get:

y= 3(x^2 + 1)^2 ----> (3x^2 + 3)^2

Do I plug this in the y-y = m(x-x) or something else?

Thanks!

 

[stapel]

Note: 3(x² + 1)² = 3(x⁴ + 2x² + 1) = 3x⁴ + 6x² + 3, not 9x⁴ + 18x² + 9 = (3x² + 3)².

Note: y - y = 0 and x - x = 0. Do you perhaps mean y - y₁ = m(x - x₁)?

If y = (x² + 1)³, then y cannot also equal 3(x² + 1)². Do you perhaps mean this second expression to be dy/dx?

Once you get dy/dx, you need to find the value at x = 1. To do this, evaluate dy/dx at x = 1; that is, plug "1" in for "x", and simplify. Since the derivative is the slope at that point, this value gives you the slope "m".

They have given you the point; you should be able to obtain the slope. Then plug x₁, y₁, and m into the point-slope form, and simplify to get the tangent-line equation.

Eliz.