How do you implicitly differentiate #x^2 y^3 − xy = 10#?

Question

William Abdalla · Accepted Answer

You can do it like this

#x^2y^3-xy=10# Differentiate both sides with respect to #x#: #D(x^2y^3-xy)=D(10)# Then apply the product rule and the chain rule: #x^(2)3y^2y'+y^(3)2x-(xy'+y)=0# #3x^(2)y^(2)y'+2y^(3)x-xy'-y=0# #:.y'(3x^2y^2-x)=y-2y^3x# #:.y'(3x^2y^2-x)=y(1-2xy^2)# #:.y'=(y(1-2xy^2))/((3x^2y^2-x))# #y'=(y(1-2xy^2))/(x(3xy^2-1)#

Charles Anctil · Answer

undefined To implicitly differentiate the equation $x^2y^3 - xy = 10$, follow these steps:

1. Differentiate each term of the equation with respect to $x$.
2. Apply the product rule and chain rule when necessary.
3. Solve for $\frac{dy}{dx}$.

Differentiating each term:
$$ \frac{d}{dx}[x^2y^3] - \frac{d}{dx}[xy] = \frac{d}{dx}[10] $$

Apply the product rule and chain rule:
$$ 2xy^3 + x^2(3y^2 \cdot \frac{dy}{dx}) - (y + x\frac{dy}{dx}) = 0 $$

Now, solve for $\frac{dy}{dx}$:
$$ 2xy^3 + 3x^2y^2 \frac{dy}{dx} - y - x\frac{dy}{dx} = 0 $$
$$ 3x^2y^2 \frac{dy}{dx} - x\frac{dy}{dx} = y - 2xy^3 $$
$$ \frac{dy}{dx}(3x^2y^2 - x) = y - 2xy^3 $$
$$ \frac{dy}{dx} = \frac{y - 2xy^3}{3x^2y^2 - x} $$

Thus, $\frac{dy}{dx} = \frac{y - 2xy^3}{3x^2y^2 - x}$.