x^2/3+y^2/3=5 find dy/dx using differentiating implicity
To find dy/dx using implicit differentiation for the given equation x^(2/3) + y^(2/3) = 5, follow these steps:
Step 1: Differentiate both sides of the equation with respect to x. Treat y as a function of x and apply the chain rule:
(d/dx) [x^(2/3) + y^(2/3)] = (d/dx) [5]
Step 2: Apply the power rule and chain rule for the left side of the equation. The derivative of x^n with respect to x is n*x^(n-1), and the derivative of y with respect to x is dy/dx:
(2/3)*x^(-1/3) + (2/3)*y^(-1/3) * (dy/dx) = 0
Step 3: Isolate the term containing dy/dx:
(2/3)*y^(-1/3) * (dy/dx) = -(2/3)*x^(-1/3)
Step 4: Solve for dy/dx by dividing both sides by (2/3)*y^(-1/3):
(dy/dx) = -[(2/3)*x^(-1/3) / (2/3)*y^(-1/3)]
Step 5: Simplify the expression by canceling out common terms:
(dy/dx) = -[(2/3)*x^(-1/3) / (2/3)*y^(-1/3)] = -((x^(-1/3)) / (y^(-1/3)))
Step 6: Rewrite the expression using positive exponents:
(dy/dx) = -((y^(1/3)) / (x^(1/3))
Therefore, the derivative dy/dx for the equation x^(2/3) + y^(2/3) = 5 is dy/dx = - ((y^(1/3)) / (x^(1/3))).