Need help with my math pls!

Question

5.Find the points at which the graph of the equation has a vertical or horizontal tangent line. 6x^2+7y^2+48x+42+8=0 

a.There are no horizontal or vertical tangent lines. 
b.There is a vertical tangent at but no horizontal tangents. 
c.There is a horizontal tangent at and a vertical tangent at . 
d.There is a horizontal tangent at but no vertical tangents. 
e.There is a horizontal tangent at and a vertical tangent at . 

6Find the slope of the tangent line (4-x)y^2=x^3 at the given point (2,2) Round your answer to two decimal places. 
a.1.00 
b.1.67 
c.0.67 
d.3.00 
e.2.00 
7.Find d2y/dx2 in terms of x and y. x^2+y^2=5 
a.(-x/y)^2 
b.-5/y^20 
c.(-y/x)^2 

8.Find by implicit differentiation given that x^6/7+y^6/7=3 
a.dy/dx= -7 square root y/x 
b.dy/dx=-7 square root x/y 
c.dy/dx=-7 square root xy 
d.dy/dx=7 square root y/x 
e.dy/dx=7 square root xy 

Answer 1

(6).

The equation is (4-x)y² = x³ and the point is (2, 2).

Differentiate with respect to x.

2(4-x)yy' - y² = 3x²

2(4-x)yy' = 3x² + y²

y' = [3x² + y²]/2(4-x)y

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

m = y' = [3*2² + 2²]/2(4-2)2 = 16/8 = 2.

The option e is correct.

 

Answer 2

  

(7).

The equation is x² + y² = 5.

Differentiate with respect to x.

2x + 2yy' = 0

x + yy' = 0

y' = -(x/y)

Again differentiate with respect to x.

y'' = - [y - xy']/y²

y'' = - [y - x{-(x/y)}]/y²    [Since y' = -(x/y)]

y'' = - [y² + x²]/y³

y'' = - 5/y³                   [Since x² + y² = 5]

None of these options are correct.

  

Answer 3

(8).

The equation is x^6/7 + y^6/7 = 3.

Differentiate with respect to x.

(6/7)x^{6/7 - 1} + (6/7)y^{6/7 - 1}y' = 0

x^-1/7 + y^-1/7y' = 0

y^-1/7y' = - x^-1/7

y' = - (x/y)^-1/7

y' = - (y/x)^1/7

The option a is correct.

Answer 4

(5).

The equation is 6x² + 7y² + 48x + 42y + 8 = 0.

Differentiate with respect to x.

12x + 14yy' + 48 + 42y' = 0

14yy' + 42y' = - (12x + 48)

(14y + 42)y' = - (12x + 48)

y' = - (12x + 48)/(14y + 42)

y' = - (6x + 24)/(7y + 21)

The slope of vertical tangent line is 1/0,

Therefore, - (6x + 24)/(7y + 21) = 1/0

7y + 21 = 0

y = -3

Substitute y = -3 in the original equation.

6x² + 7(-3)² + 48x + 42(-3) + 8 = 0

6x² + 48x - 55 = 0

x² + 8x - 55/6 = 0

(x² + 8x + 16) - 16 - 55/6 = 0

(x + 4)² - 151/6 = 0

(x + 4)² = 151/6

x + 4 = ± √(151/6)

x = - 4 ± √(151/6)

x = - 4 + √(151/6) and x = - 4 - √(151/6)

x = 1.02 and x = - 9.02

The points are (1.02, -3) and (-9.02, -3).

The slope of horizontal tangent line is 0,

Therefore, - (6x + 24)/(7y + 21) = 0

6x + 24 = 0

x = -4

Substitute x = -4 in the original equation.

6(-4)² + 7y² + 48(-4) + 42y + 8 = 0

7y² + 42y - 88 = 0

y² + 6y - 88/7 = 0

(y² + 6y + 9) - 9 - 88/7 = 0

(y + 3)² - 151/7 = 0

(y + 3)² = 151/7

y + 3 = ± √(151/7)

y = - 3 ± √(151/7)

y = - 3 + √(151/7) and y = - 3 - √(151/7)

y = 1.64 and y = - 7.64

The points are (-4, 1.64) and (-4, -7.64).