tangent line to a curve?

Question

Find an equation of the tangent line to the curve at the point (2, 0).
y = ln (x^3 - 7)

i think the derivative is 1/x^3-7 *3x^2

Answer 1

then plug in 2 to the derivative... that is the slope

1/(8-7) * 3*4 = 12

thus
tangent line:
y = 12(x-2)


§

Answer 2

That is in fact the derivitive. To find the tangent line, you must first find the slope at that point. The slope of the line at that point is the derivitive at that point, so simply plug 2 into the x's in your derivitive equation:

3(4) / (8-7) = 12

12 is the derivitive there, as well as the slope at the point (2,0)

if you use the form a a straight line y = mx + b, then m is the slope, or in this case 12. From there it is muce easier. Since you know y = 0 when x = 2, simply plug these in to figure out b:

0 = (12)(2) + b

b = -24.

The answer: y = 12x - 24.

Answer 3

Given y = ln (x^3 - 7 )
Therefore y1 = (3.x^2) / ( x^3 - 7 )

(y1) x = 2 = 3.2^2 / (2^3 - 7 )
= 12

So the required Tangent can be written as

y = 12x + c where c is y-intercept. ........... Eqn 1

But this tangent passes through (2,0)

Therefore substituting in Eqn 1 we get

0 = 12(2) + c Whence C = - 24

Hence the required Eqn comes out to be -

12x - y = 24 ANSWER

Answer 4

y = ln (x^3 - 7)
=> dy/dx = 3x^2 / (x^3 - 7)
=> dy/dx ( at (2, 0)) = 3(2)^2/(2^3 - 7) = 12

Hence equation of tangent at (2, 0) is
y -0 = 12(x - 2)
=> 12x - y = 24.

Answer 5

y = ln u
dy/du = 1 / u
u = x³ - 7
du/dx = 3x²
dy/dx = (dy/du)(du/dx)
dy/dx = (1/u) (3x²)
dy/dx = (1 / (x³ - 7) ) (3 x ² )
Which agrees with your solution.