Derivatives: Logarithmic Function?

Question

Find an equation of the tangent line to the curve
y=(3 x2+3 )ln(3 x2-2 )+5
when x= 1.

Solution: We use the fact that the tangent line to the curve
y=f(x) when x=x0 is given by
y=f '(x0)(x-x0)+f(x0).

For
y=f(x)=(3 x2+3 )ln(3 x2-2 )+5

and x0= 1
we have that f(1)=_________ and f '(1)= ___________ .

An equation of the tangent line to the given curve when x0= 1 is y=_____________ .

Answer 1

fist take the derivative of the function
y' = (6x)*(ln(3x^2 - 2)) + (3x^2 + 3)*(6x)/(3x^2 - 2)
(y - y0)/(x - x0) = m , is the equation of a line with constant gradient.
Which simplifies to your equation as m is equal to the gradient (y'(x0)) at that point.
Sub in your values
(y - 5)/(x - 1) = 36
y = 36x - 35

Answer 2

y = (3x^2 + 3)ln(3x^2 - 2) + 5
y' = 6x(3x^2 + 3)/(3x^2 - 2) + 6xln(3x^2 - 2)

y(1) = (3(1)^2 + 3)ln(3(1)^2 - 2) + 5 = 5
y'(1) = 6(3(1)^2 + 3)/(3(1)^2 - 2) + 6ln(3(1)^2 - 2)
y'(1) = 6(3 + 3)/1 = 36

equation of the tangent line at x = 1 is
y = 36(x - 1) + 5
or
y = 36x - 31