Use implicit differentiation to find an equation of the tangent line to the curve x^(2/3) + y^(2/3) =?

Question

Use implicit differentiation to find an equation of the tangent line to the curve x^(2/3) + y^(2/3) = 6 at the point ( -5*sqrt(5), 1)

Give your answer in y = mx + b

Answer 1

The slope of the curve is given by the implicit differentiation.
(d/dx)(x^(2/3) + y^(2/3)) = (d/dx)(6)
=> (2/3)x^(2/3 - 1) + (2/3)y^(2/3 - 1) (dy/dx) = 0
=> (2/3)x^(-1/3) + (2/3)y^(-1/3) (dy/dx) = 0
=> (2/3)y^(-1/3) (dy/dx) = -(2/3)x^(-1/3)
=> dy/dx = -(2/3)x^(-1/3) * (3/2)y^(1/3)
=> dy/dx = -(y/x)^(1/3)

The slope, then, is dy/dx(-5*sqrt(5),1) = -(1/-5*sqrt(5))^(1/3)
= 1/(5*sqrt(5))^(1/3). Let us denote this by m.

Now the tangent line would be
y = mx + b, which passes through (-5*sqrt(5), 1).
1 = m(-5*sqrt(5)) + b, i.e. b = 1 + 5*sqrt(5)m
= 1 + (5*sqrt(5))^(2/3).

Thus, the tangent line is
y = x/(5*sqrt(5))^(1/3) + (1 + (5*sqrt(5))^(2/3)).

There might be some mistake in computation since
I haven't double-checked, but the idea is there.

Answer 2

Both of the previous people correctly got dy/dx as a function but both put in x = 1, y = 2 too quickly and made mistakes. Big D's last three lines should be (-2 -4 -1)/(2 - 4) = -7/(-2) = 7/2 so tangent is y - 2 = (7/2)(x - 1) ---> y = (7x/2) - (3/2) Hk also should have got 7/2 for gradient at that point. I have noticed before that Puggy can't do implicit differentiation correctly. I agree with other two on the form of dy/dx.

Answer 3

Implicitly differentiating x^2/3 + y^2/3 = 6 we get
(2/3)x^(-1/3)dx + (2/3)y^(-1/3) dy = 0

Simplifying
dy/dx = -(y/x)^(1/3)

Thus, the slope at (-5*sqrt(5), 1) is [1/(5sqrt(5)]^(1/3)

y - 1 = [1/(5sqrt(5)]^(1/3) (x + 5 sqrt(5))
y = [1/(5sqrt(5)]^(1/3)]x + [(5 sqrt(5))/(5 sqrt(5))^(1/3)] + 1

The constant term on the RHS is 1 +(5 sqrt(5))^(2/3)
The tangent is

y = [1/(5sqrt(5)]^(1/3)]x + (1 + (5 sqrt(5))^(2/3))

Answer 4

x^(2/3) + y^(2/3) = 6 at the point ( -5*sqrt(5), 1)
(2/3) x'/x^(1/3) + (2/3) y'/y^(1/3) = 0
dy/dx = -(y/x)^(1/3) = (1/(5sqrt(5)) )^(1/3) = 1/sqrt(5)

y-1 = (1/sqrt(5))( x+5sqrt(5))
y = (1/sqrt(5))x + 6

Answer 5

x^(2/3) + y^(2/3) = 6
(2/3)x^(-1/3) + (2/3)y^(-1/3)y' = 0
(2/3)y^(-1/3)y' = -(2/3)x^(-1/3)
y' = -[x^(-1/3)]/y^(-1/3)
y' = -(y/x)^(1/3)

at (-5√5, 1)

y' = -[ -1/(5√5)]^(1/3)
y' = (1/√125)^(1/3)
y' = 1/√5

so line is

y - 1 = (1/√5)(x + 5√5)
y = (1/√5)x + 5 + 1
y = (1/√5)x + 6