Can help me with DIFFERENTIATION?

Question

1) differentiate with respect to X.

2x^2 + 8√x + (x^2 + 3x) / (x^2)

2)
the curve c has equatio3n y=7-4x-4/x.
x>0.
the point p on c has x-coordinate equal to 2.

a) show that the equation of the tangent c at the point P is y=-3x-3.
b) Find an equation of the normal to c at the point p.
The tangent at p meets the x-axis at A and the normal at P meets the x-axis at B.
C) Find the area of the triangle APB

Answer 1

(1).

The expression is 2x² + 8√x + (x² + 3x)/x².

= 2x² + 8√x + (x²)/x² + (3x)/x².

= 2x² + 8√x + 1 + 3/x

Differeniate with respect ot x.

= 4x + 4/√x - 3/x².

The differentiate of 2x² + 8√x + (x² + 3x)/x² = 4x + 4/√x - 3/x².

Answer 2

2(a).

The curve equation y = 7 - 4x - 4/x and the x-coordinate of the point P is 2.

To find the y-coordinate of the point P, substitute x = 2 in the original curve equation.

y = 7 - 4(2) - 4/(2) = 7 - 8 - 2 = -3.

Differentiate with respect to x to the curve equation.

y' = - 4 + 4/x².

Find slope of tangent line at the point P(2, -3).

m = - 4 + 4/(2)² = -4 + 1 = -3.

Point-slope form of line equation : y - y₁ = m(x - x₁), where m = slope and (x₁, y₁) = point.

y - (-3) = (-3)(x - 2)

y + 3 = -3x + 6

y = -3x + 3.

The tangent line equation is y = -3x + 3.

Answer 3

2(b).

The curve equation y = 7 - 4x - 4/x and the x-coordinate of the point P is 2.

To find the y-coordinate of the point P, substitute x = 2 in the original curve equation.

y = 7 - 4(2) - 4/(2) = 7 - 8 - 2 = -3.

Differentiate with respect to x to the curve equation.

y' = - 4 + 4/x².

Find slope of tangent line at the point P(2, -3).

m = - 4 + 4/(2)² = -4 + 1 = -3.

The slope of normal line at the point P(2, -3) = - 1/m = -1/(-3) = 1/3.

Point-slope form of line equation : y - y₁ = m(x - x₁), where m = slope and (x₁, y₁) = point.

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

3(y + 3) = x - 2

3y + 9 = x - 2

3y = x - 11

y = (1/3)x - (11/3).

The normal line equation is y = (1/3)x - (11/3).