8. Remember that the derivative, dy/dx gives you the slope of the tangent line for any given value of x. So let's find dy/dx (or y', means the same thing), we will have to do this using implicit differentiation, i hope you know it.

x^3 + y^3 - 3xy = 17

=> 3x^2 + 3y^2 y' - 3y - 3x y' = 0 ....now we group the y' together

=> 3y^2 y' - 3x y' = -3x^2 + 3y

=> y'(3y^2 - 3x) = 3y -3x^2

=> y' = (3y - 3x^2)/(3y^2 - 3x)

now at (2,3), y' = [3(3) - 3(2)^2]/[3(3)^2 - 3(2)] = (-3)/(21) = -1/7

now recall that the equation of a line is of the form y = mx + b, where m is the slope given by the derivative and b is the y-intercept. we will use the point-slope form to find the equation of the tangent line.

y - y1 = m(x - x1)

=> y - 3 = (-1/7)(x - 2)

=> y = (-1/7)x + 2/7 + 3

=> y = (-1/7)x + 23/7 ............equation of the tangent line.

9. Given the curve y = kx^2 + 5 and the eqution of the tangent line 4x + y = 9

First let's see where these lines intersect. we have:

y = -4x + 9

y = kx^2 + 5

=> kx^2 + 5 = -4x + 9

=> kx^2 + 4x - 4 = 0

=> x = [-4 +/- sqrt(16 + 16k)]/2k ............by the quadratic formula

=> x = [-4 +/- -4sqrt(1 + k)]/2k

=> x = [-2 +/- -2sqrt(1 + k)]/k

now y = kx^2 + 5

=> y' = 2kx

since the slope of the given tangent line is -4, we want y' = 2kx = -4

=> 2kx = -4

=> kx = -2

=> x = -2/k

now substitute this value for x in the formula we made above.

x = [-2 +/- -2sqrt(1 + k)]/k

=> -2/k = [-2 +/- -2sqrt(1 + k)]/k

=> -2 = -2 +/- -2sqrt(1 + k)

=> 0 = +/- -2sqrt(1 + k)

=> 0 = +/- sqrt(1 + k)

=> 0 = 1 + k ................squared both sides

=> k = -1

10. f(x) = sqrt(2 - x)

they want us to find the derivative of f(x) using first principles. this means they want us to find it by the limit definition:

using f ' (x) = limit{h-->0} [f(x + h) - f(x)]/h

=> f ' (x) = limit{h-->0} [sqrt(2 - x - h) - sqrt(2 - x)]/h ...now we abandon our mathematical training for a moment and irrationalize the quotient.

=> f ' (x) = limit{h-->0} [sqrt(2 - x - h) + sqrt(2 - x)]/[sqrt(2 - x - h) + sqrt(2 - x)] * [sqrt(2 - x - h) - sqrt(2 - x)]/h

=> f ' (x) = limit{h-->0} [2 - h - x - (2 - x)]/h[sqrt(2 - x - h) + sqrt(2 - x)]

=> f ' (x) = limit{h-->0} -h/h[sqrt(2 - x - h) + sqrt(2 - x)]

=> f ' (x) = limit{h-->0} -1/[sqrt(2 - x - h) + sqrt(2 - x)]

finally we can take the limit without worrying about the denominator going to zero.

=> f ' (x) = -1/[sqrt(2 - x) + sqrt(2 - x)]

=> f ' (x) = -1/[2*sqrt(2 - x)]

=> f ' (x) = -(1/2)(2 - x)^(-1/2) ..........as expected

if there's anything you don't understand about any of the questions, dont hesitate to ask