# How do you implicitly differentiate x^3 y-3xy^2 at the point (1,-10)?

##### 1 Answer
Aug 14, 2015

Refer Explanation

#### Explanation:

${x}^{3}$y - 3${y}^{2}$ = 0

To differentiate ${x}^{3}$y use product rule

${x}^{3}$.$\frac{\mathrm{dy}}{\mathrm{dx}}$ + y.3${x}^{2}$ - 6y$\frac{\mathrm{dy}}{\mathrm{dx}}$ = 0

Rewrite it -

${x}^{3}$.$\frac{\mathrm{dy}}{\mathrm{dx}}$ + 3${x}^{2}$y - 6y$\frac{\mathrm{dy}}{\mathrm{dx}}$ = 0

Keep $\frac{\mathrm{dy}}{\mathrm{dx}}$ terms on the left hand side

${x}^{3}$.$\frac{\mathrm{dy}}{\mathrm{dx}}$ - 6y$\frac{\mathrm{dy}}{\mathrm{dx}}$ = - 3${x}^{2}$y

Take $\frac{\mathrm{dy}}{\mathrm{dx}}$ as common factor

$\frac{\mathrm{dy}}{\mathrm{dx}}$(${x}^{3}$ - 6y) = - 3${x}^{2}$y
Solve it for $\frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}}$ = $\frac{- 3 {x}^{2} y}{{x}^{3} - 6 y}$

At x = 1 and Y = - 10

$\frac{\mathrm{dy}}{\mathrm{dx}}$ = $\frac{- 3 \left({1}^{2}\right) \left(- 10\right)}{{1}^{3} - 6 \left(- 10\right)}$ = $\frac{30}{61}$