Question

Find the equation of the line tangent to the graph of y=(x^2 - x)ln(6x) at x=2. Approximate to the nearest thousandth.

Answer 1

`y = 8.4547x - 11.9396`

Explanation:

We are given the following information in the question:

`y = (x^2-x)ln(6x)`

Differentiating y with respect to x:

`\displaystyle(dy)/(dx) = (x^2-x)'ln(6x) + (x^2-x)(ln(6x))'\\=(2x-1)ln(6x) + (x^2-x)(6)/(6x)\\\\= (2x-1)ln(6x) + (x-1)`

At x =2

`\displayatyle(dy)/(dx)\Bigr|_{\substack{x=2} }= (4-1)ln(12) + (2-1) = 8.4547`

`y(2) = (2^2-2)ln(12) = 4.9698`

Equation of tangent:

`(y-y_0) = \displaystyle(dy)/(dx)(x-x_0)`

Putting the values:

`(y-y(2)) = \displaystyle(dy)/(dx)\Bigr|_{\substack{x=2} }(x-2)\\\\y - 4.9698 = 8.4547(x-2)\\y - 4.9698 = 8.4547x - 16.9094\\y = 8.4547x - 11.9396`

The above equation is the required equation of the tangent.