Consider the curve given by the equation (2y+1)^3 − 24x = −3.

(a) Show that dy/dx = 4/(2y+1)^2.

(b) Write an equation for the line tangent to the curve at the point (−1,−2).

(c) Evaluate d2y/dx2 at the point (−1,−2).

(d) The point (16,0) is on the curve. Find the value of (y−1)′(0).

(a) dy/dx = 4/(2y+1)^2.

(b) y = 4/9 x - 14/9

(c) d2y/dx2 = -64/243

Step-by-step explanation:

You have the following equation

  (1)

(a) You first derivative implicitly the equation (1) respect to x:

next, you solve the last result for dy/dx:

(2)

(b) The equation for the tangent line is given by:

   (3)

with yo = -2 and xo = -1

To find the slope m you use the result of the equation (2), because dy/dx evaluated in (-1,-2) is the slope at such point:

m =

Hence, by replacing in the equation (3) you obtain:

hence, the equation for the tangent line is y = 4/9 x - 14/9

(c) To find d2y/dx2 you derivative the result obtain in the equation (2):

    (4)

the second derivative for the point (-1,-2) is obtained by replacing y=-2 and dy/dx=m=4/9 in the equation (4):

hence, d2y/dx2 evaluated in (-1,-2) is -64/243