Calculus Help? Derivatives and Rates of Change.

There were a few problems I was stuck on. Any help would be much appreciated!

Find an equation of the tangent line to the curve at the given point.
5. y= (x-1)/(x-2) ; (3,2)
7. y= sqrt(x) ; 1,1

also,
if f(x) = 3x^2 - 5x, find f '(2) and use it to find an equation of the tangent line to the parabola y= 3x^2 - 5x at the point (2,2)

Again any help at all would be much appreciateddd! Thank you.

These problems are all solved the same way. The derivative gives the slope of the tangent, then use point/slope to write the equation.
5. Y' =[(x-2)(1)-(x-1)(1)]/(x-2)^2

Hi, again ! This is the quotient rule. The derivative of a rational function is given by

[Denominator * der of the numerator - numerator * der of the denom.]/(denom^2).

Or d(u/v)= (vdu-udv)/v^2

= -1/(x-2)^2
At (3,2), plug in 3 for x. M=-1
Then the tangent line is y-2= -1(x-3)

7. Y' =(1/2)x^(-1/2)
At (1,1), m= 1/2
so y-1 = (1/2)(x-1)

Then for the parabola, y' = 6x-5 and f'(2) =7 which is the slope
At (2,2) the tangent line is y-2=7(x-2)

f'(x) = 6x - 5
so f'(2)= 6(2)-5 = 7