Question

Differentiate x2/3 - y2/3 = 6 with respect to x and evaluate the derivative at (8, 1).

Answer 1

Implicit Differentiation

Explanation:

Given:

`x^{(2)/(3)} -y^{(2)/(3)} = 6\\`

Recall:

`x^{(2)/(3)} -y^{(2)/(3)} = 6 \\ \frac{\text{d}}{\text{d}x}(x^{(2)/(3)} -y^{(2)/(3)}) = \frac{\text{d}}{\text{d}x}(6) \\ \frac{\text{d}}{\text{d}x}(x^{(2)/(3)}) -\frac{\text{d}}{\text{d}x}(y^{(2)/(3)}) = 0 \\ (2)/(3)x^{(2)/(3) -1} -(2)/(3)y^{(2)/(3) -1}\cdot \frac{\text{d}y}{\text{d}x} = 0 \\ (2)/(3)x^{-(1)/(3)} - (2)/(3)y^{-(1)/(3)} \cdot \frac{\text{d}y}{\text{d}x} = 0 \\ \frac{2}{3\sqrt[3]{x}} -\frac{2}{3\sqrt[3]{y}}\cdot \frac{\text{d}y}{\text{d}x} = 0 \\ -\frac{2}{3\sqrt[3]{y}}\cdot \frac{\text{d}y}{\text{d}x} = -\frac{2}{3\sqrt[3]{x}} \\ -2 \cdot \frac{\text{d}y}{\text{d}x} = -\frac{6\sqrt[3]{y}}{3\sqrt[3]{x}} \\ \frac{\text{d}y}{\text{d}x} = -\frac{6\sqrt[3]{y}}{-6\sqrt[3]{x}} \\ \frac{\text{d}y}{\text{d}x} = \frac{\sqrt[3]{y}}{\sqrt[3]{x}}`