how to solve

Question

1. Given: y=e^3x.cos3x Differentiate by using the product rule. 2. Differentiate the following with respect to x: y=2.3^5x - 4cos 2x - 5lnx - 3x^-1 3. Given: y=x^3 - 11x^2 + 32x -28 Determine, using differentiation, the maximum and the minimum turning points. Also determine the point of inflection by using the second derivative.

Answer 1

1.

y= e^3x.cos3x

Differentiate with respect to x using product rule.

(uv)' = u v' + v u'

u = e^3x => u' =e^3x* (3x)' = 3 e^3x

v = cos (3x) => v' =- sin(3x) * (3x)' = - 3 sin(3x)

y ' = e^3x (cos (3x))' + cos (3x) (e^3x)'

   = - 3e^3x sin(3x) + 3 cos (3x) (e^3x)

   = 3e^3x [ cos(3x) -  sin (3x)]

Answer 2

3)

f(x) = x^3 - 11x^2 + 32x -28

Take derivative both sides with respected to x.
f ' (x) = 3(x²) - 11(2x) + 32
        = 3x² - 22x + 32.
Again derivative to each side with respect to x.
f " (x) = 6(x) - 22.
         = 6x - 22.
To find the turning points., to make the first derivative equal to zero or f ' (x) = 0.
3x² - 22x + 32 = 0
3x² - 6x - 16x + 32  = 0.
3x(x - 2) - 16(x - 2) = 0.
(x - 2)(3x - 16) = 0.
3x - 16= 0 and x - 2 = 0.
x = 16/3 and x = 2.

f " (c) > 0 (positive) ------> minimum point.
f " (c) < 0 (negative) ------> maximum point.

f " (x) = 6x - 22.
At x = 16/3

f " (16/3) = 6(16/3) - 22 = 10>0

At x = 16/3, f(x) is minimum.

Find the the minimum value to substitute x = 16/3 in f(x).

f(16/3) =(16/3)^3 - 11(16/3)^2 + 32(16/3) -28

          = (4096/27) - (2816/9) +(512/3) - 28

          = -18.52

At x = 2

f " (2) = 6(2) - 22 = -10<0

At x = 2, f(x) is maximum.

Find the the maximum value to substitute x = 2 in f(x).

f(-1) = (2)^3 - 11(2)^2 + 32(2) -28

          = 8 - 44 +64 - 28

          = 0

To find inflection point .

f " (x) = 0

6x - 22 = 0

x = 22/6

f(x) = (22/6)^3 - 11(22/6)^2 + 32(22/6) -28

     = - 5.6

The inflection point is  (22/6, -5.6)