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

Answer:

$y = 8.4547x - 11.9396$    

Step-by-step explanation:

We are given the following information in the question:

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

Differentiating y with respect to x:

$\displaystyle\frac{dy}{dx} = (x^2-x)'ln(6x) + (x^2-x)(ln(6x))'\\=(2x-1)ln(6x) + (x^2-x)\frac{6}{6x}\\\\= (2x-1)ln(6x) + (x-1)$

At x =2

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

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

Equation of tangent:

$(y-y_0) = \displaystyle\frac{dy}{dx}(x-x_0)$

Putting the values:

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

The above equation is the required equation of the tangent.