How do you implicitly differentiate #y sin(x^2) = x sin (y^2)#?

Question

Paisley Johnson · Accepted Answer

In deriving remember that #y# is a fnction of #x#:

Isaiah Allen · Answer

undefined To implicitly differentiate $ y \sin(x^2) = x \sin(y^2) $, follow these steps:

1. Differentiate both sides of the equation with respect to $x$.
2. Apply the product rule and chain rule as needed.
3. Solve for $\frac{dy}{dx}$ to find the implicit derivative.

Here are the steps in detail:

Given the equation: $ y \sin(x^2) = x \sin(y^2) $

1. Differentiate both sides with respect to $x$:
$$ \frac{d}{dx}(y \sin(x^2)) = \frac{d}{dx}(x \sin(y^2)) $$

2. Apply the product rule and chain rule:
$$ \frac{dy}{dx} \sin(x^2) + y \frac{d}{dx}(\sin(x^2)) = \sin(y^2) + x \cos(y^2) \frac{dy}{dx} \frac{d}{dx}(y^2) $$

3. Simplify and solve for $\frac{dy}{dx}$:
$$ \frac{dy}{dx} \sin(x^2) + y \cdot 2x \cos(x^2) = \sin(y^2) + 2xy \cos(y^2) \frac{dy}{dx} $$

$$ \frac{dy}{dx} \sin(x^2) - 2xy \cos(y^2) \frac{dy}{dx} = \sin(y^2) - y \cdot 2x \cos(x^2) $$

$$ \frac{dy}{dx}(\sin(x^2) - 2xy \cos(y^2)) = \sin(y^2) - y \cdot 2x \cos(x^2) $$

$$ \frac{dy}{dx} = \frac{\sin(y^2) - y \cdot 2x \cos(x^2)}{\sin(x^2) - 2xy \cos(y^2)} $$

That's the implicit derivative of the given equation.