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Question

1. Use implicit differentiation to find an equation of the tangent line to the ellipse x^2/2+y^2/98=1 at (1,7) 

a.-8x+9 
b.-2x+9 
c.-13x+9 
d.-3x+11 
e.-5x+11 

2.Find dy/dx by implicit differentiation. 
x^3/4+y^4/5=4 

3.Find by implicit differentiation given that tan (2x+y)=2x Use the original equation to simplify your answer 

a.-2x/x^2-1 
b.-2x^2/x^2+1 
c.2x^2/x^2-1 
d.2x/x^2+1 
e.2x^2/x^2+1 

4.Find an equation of the tangent line to the graph of the function given below at the given point. 8x^2-4xy+6y^2-10=0 , (-1,-1) 

a.3.50x-2.50 
b.3.50x+2.50 
c.1.50x-2.50 
d-1.50x+2.50 
e.-1.50x-2.50 

Answer 1

(1).

The ellipse equation is x²/2 + y²/98 = 1 and the point is (1, 7).

Differentiate with respect to x.

x + (y/49)y' = 0

y' = - 49x/y

Substitute (x, y) = (1, 7) in the above equation.

m = y' = - 49(1/7) = -7.

Substitute (x₁, y₁) = (1, 7) m = -7 in the point-slope form of equation: y - y₁ = m(x - x₁).

y - 7 = -7(x - 1)

y - 7 = -7x + 7

y = -7x + 14.

The tangent line equation is y = -7x + 14.

None of these options are correct.

Answer 2

(2).

The equation is x³/4 + y⁴/5 = 4.

Differentiate with respect to x.

(d/dx)[x³/4 + y⁴/5] = (d/dx)(4)

(d/dx)(x³/4) + (d/dx)(y⁴/5) = 0

(3/4)x² + (4/5)y³y' = 0

4y³y'/5 = - 3x²/4

y' = (-3x²/4)(5/4y³)

y' = -15x²/16y³

Answer 3

(3).

The equation is tan(2x + y) = 2x.

Differentiate with respect to x.

(d/dx)[tan(2x + y)] = (d/dx)(2x)

sec²(2x + y)*(d/dx)(2x + y) = 2

sec²(2x + y)(2 + y') = 2

2 sec²(2x + y) + y' sec²(2x + y) = 2

y' sec²(2x + y) = 2 - 2 sec²(2x + y)

y' sec²(2x + y) = 2[1 - sec²(2x + y)]

y' = 2{1 - sec²(2x + y)}/[sec²(2x + y)]

y' = 2[- tan²(2x + y)/ {1 + tan²(2x + y)}]        [Since sec²(θ) - tan²(θ) = 1 ]

y' = 2[- (2x)²/ {1 + (2x)²}]                               [Since tan(2x + y) = 2x ]

y' = -8x²/(1 + 4x²)

None of these options are correct.

Answer 4

(4).

The function is 8x² - 4xy + 6y² - 10 = 0 and the point is (-1, -1).

Differentiate with respect to x.

16x - 4[xy' + y] + 12yy' = 0

4x - xy' - y + 3yy' = 0

[- x + 3y]y' = y - 4x.

y' = (y - 4x)/(- x + 3y).

Substitute (x, y) = (-1, -1) in the above equation.

m = y' = (- 1 + 4)/(1 - 3) = - 3/2 = - 1.5.

Substitute (x₁, y₁) = (-1, -1) m = -1.5 in the point-slope form of equation: y - y₁ = m(x - x₁).

y + 1 = -1.5(x + 1)

y + 1 = -1.5x - 1.5

y = -1.5x - 2.5

The tangent line equation is y = -1.5x - 2.5.

The option e is correct.