Equation of Tangent and ln

[Jason76]

Find equation of the tangent line for \(\displaystyle (3,0)\)

\(\displaystyle y = \ln(x^{2} - 3x + 1)\)

\(\displaystyle y' = \ln(2x - 3)\)

\(\displaystyle y' = \dfrac{2}{2x - 3}\)

\(\displaystyle y'(0) = \dfrac{2}{2(3) - 3} = \dfrac{2}{3}\)

\(\displaystyle y - 0 = \dfrac{2}{3}(x - 3)\)

\(\displaystyle y - 0 = \dfrac{2}{3}x - 2\)

\(\displaystyle y = \dfrac{2}{3}x - 2\)

 

[HallsofIvy]

Sorry but you seem to be completely confused here.

Find equation of the tangent line for \(\displaystyle (3,0)\)

\(\displaystyle y = \ln(x^{2} - 3x + 1)\)

\(\displaystyle y' = \ln(2x - 3)\)

No, you, can't just differentiate inside a function

\(\displaystyle y' = \dfrac{2}{2x - 3}\)

And now, having differentiated just part you differentiate the new function?

You are trying to use the "chain rule" with y= ln(u) and \(\displaystyle u= x^2-3x+ 1\). The derivative of ln(u) is
(1/|u|)(du/dx) (notice the absolute value). \(\displaystyle du/dx= d(x^2- 3x+ 1)/dx= 2x- 3\) so
the derivative of \(\displaystyle ln(x^2- 3x+ 1)\) is \(\displaystyle (2x-3)/|x^2- 3x+ 1|\)

\(\displaystyle y'(0) = \dfrac{2}{2(3) - 3} = \dfrac{2}{3}\)

y'(0)? (3, 0) means x= 3, y= 0. You want to find the tangent line at the point (3, 0) on the graph so you should be evaluating \(\displaystyle y'= (2x-3)/|x^2- 3x+ 1|\) at x= 3, not 0.

\(\displaystyle y - 0 = \dfrac{2}{3}(x - 3)\)

\(\displaystyle y - 0 = \dfrac{2}{3}x - 2\)

\(\displaystyle y = \dfrac{2}{3}x - 2\)

 

[Jason76]

ok i see the error

 

[fcabanski]

This would all be nice if derivative of ln(u) = 1/|u|, but it doesn't.

Derivative of ln(u) = 1/u.