find the slop of tangent line to the following curves

Question

1. y=x^10+1+root square (1-x),

p(0,2)

2. x^3-2x^2y+y^3-5=0,

p(1,2)

 

Answer 1

1).

The curve is y = x^10 + 1 + √(1 - x) and the point is p(0, 2)

We can use differentiation to find y '.

The  total derivation of y = x^10 + 1 + √(1 - x) is

y' = 10x^9 + (1 / 2)(1 - x)^(1/2 - 1)(- 1)

 y' = 10x^9 - 1 / 2(1 - x)^(- 1/2)

Substitute the values of (x, y) = (0, 2) in the above equation.

 y' = 10(0)^9 - 1 / 2(1 - 0)^(- 1/2)

 y' = 0 - 1 / 2  = - 1 / 2.

This is the slope (m ) of the tangent line to the curve at p(0, 2).

2).

The curve is x^3 - 2x^2y + y^3 - 5 = 0 and the point is p(1,2).

We can use differentiation to find y '.

The  total derivation of x^3 - 2x^2y + y^3 - 5 = 0 is

3x^2 - 2(x^2y' + 2yx) + 3y^2y' = 0

3x^2 - 2x^2y' - 4xy + 3y^2y' = 0

y'(3y^2 - 2x^2) = 4xy - 3x^2

y' = (4xy - 3x^2) / (3y^2 - 2x^2)

Substitute the values of (x, y) = (1, 2) in the above equation.

y' = (4(1)(2) - 3(1)^2) / (3(2)^2 - 2(1)^2)

y' = (8 - 3) / (12 - 2) = 5 / 10 = 1 / 2.

This is the slope (m ) of the tangent line to the curve at p(1, 2).