Question

Use implicit differentiation to find an equation of the tangent line to the curve at the given point.

Answer 1

Differentiate both sides of

$x^{2/3} + y^{2/3} = 4$

implicite with respect to x :

$\dfrac23 x^{-1/3} + \dfrac23 y^{-1/3} \dfrac{\mathrm dy}{\mathrm dx} = 0$

Solve for the derivative dy/dx :

$x^{-1/3} + y^{-1/3} \dfrac{\mathrm dy}{\mathrm dx} = 0 \\\\ y^{-1/3} \dfrac{\mathrm dy}{\mathrm dx} = - x^{-1/3} \\\\ \dfrac{\mathrm dy}{\mathrm dx} = -\dfrac{x^{-1/3}}{y^{-1/3}} \\\\ \dfrac{\mathrm dy}{\mathrm dx} = -\left(\dfrac yx\right)^{1/3}$

Get the slope at the given point (-3√3, 1) :

$\dfrac{\mathrm dy}{\mathrm dx}(-3\sqrt3,1) = -\left(\dfrac{1}{-3\sqrt3}\right)^{1/3} = \dfrac1{(3\sqrt3)^{1/3}} = \dfrac1{\sqrt3}$

Then the equation of the tangent line to the curve through (-3√3, 1) is

$y - 1 = \dfrac1{\sqrt3}(x+3\sqrt3) \\\\ \boxed{y = \dfrac1{\sqrt3}x + 4}$