The complete question is
Use implicit differentiation to find an equation of the tangent line to the curve at the given point. [tex]x^\frac{2}{3} + y^\frac{2}{3} = 4[/tex], [tex]( -3\sqrt{3} ,1)[/tex]
Answer:
Equation of tangent is y = [tex]\frac{1}{\sqrt{3} }[/tex]x + 4
Step-by-step explanation:
We are given the equation
[tex]x^\frac{2}{3} + y^\frac{2}{3} = 4[/tex],
upon differentiating
d([tex]x^\frac{2}{3} + y^\frac{2}{3} = 4[/tex]) /dx = d(4)/dx
[tex]\frac{2}{3}x^-\frac{1}{3 } + \frac{2}{3}y^ -\frac{1}{3} dy/dx = 0[/tex]
dy/dx = [tex]-x^\frac{1}{3} /y^-\frac{1}{3}[/tex] = [tex]-y^\frac{1 }{3}/ x^\frac{1}{3}[/tex]
upon substituting the values (x, y)= [tex]( -3\sqrt{3} ,1)[/tex]
dy/dx = [tex]\frac{1}{\sqrt{3} }[/tex]
equation of the tangent
y - 1 = [tex]\frac{1}{\sqrt{3} }[/tex] ( x- (-[tex]3\sqrt{3}[/tex]))
y = [tex]\frac{1}{\sqrt{3} }[/tex]x + 4
