I need help with finding an equation of tangent to the curve at specific point

So point P lies on y=[3x+ln2x]^2

x-coordinate of P is 0.5

How to I find the tangent?

y = (3x + ln(2x) )^2

y(1/2) = (3/2 + ln(1))^2 = 2.25

y ' = 2 (3x + ln(2x)) * d/dx [ 3x + ln(2x) ]

y ' = 2 (3x + ln(2x)) * (3 + 1/x) <=== slope of curve at x

y ' (1/2) = 2 (3/2 + ln(1)) * (3 + 2) <=== slope of curve at x = 1/2

y ' (1/2) = 15

So... slope = 15 and passes through (0.5 , 2.25)

(y - 2.25) = 15 (x - 0.5)

y = 15x - 5.25

graphic check:
http://www.wolframalpha.com/input/?i=y+%…

y = (3x + ln(2x))^2

y(0.5) = (1.5 + ln 1)^2 = (1.5)^2 = 9/4

y ' = 2(3x + ln(2x))*(3 + 1/x)

y ' (0.5) = 2(1.5)(3 + 2) = 15

eqn of tangent with slope 15 and at (0.5, 9/4) is

y - 9/4 = 15(x - 1/2)

y - 9/4 = 15x - 15/2

y = 15x - 21/4

y = 15x - 5.25

Find y
use x and y in (y-a)=m(x-b) where a=y1 b=y2
differentiate using the chain rule. Remember f'(x)=1/ax if f(x)=Inax
plug this value in for m.

Bingo.

y(0.5) = 1.5^2 = 2.25
y' = 2[3x+ln2x][3+1/x] = 2(1.5)(3+2) = 15
Answer: y - 2.25 = 15(x-0.5)