# What is the implicit derivative of #1= xysin(xy) #?

Hence implict derivative is given by

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To find the implicit derivative of the equation (1 = x y \sin(xy)), we can differentiate both sides of the equation with respect to (x), treating (y) as a function of (x) using the product rule and chain rule where necessary.

Starting with the given equation:

(1 = x y \sin(xy))

Differentiating both sides with respect to (x):

(\frac{d}{dx} (1) = \frac{d}{dx} (x y \sin(xy)))

On the left side, the derivative of a constant (1) is zero.

On the right side, we use the product rule:

(\frac{d}{dx}(x y \sin(xy)) = x \frac{dy}{dx} \sin(xy) + y \frac{dx}{dx} \sin(xy) + xy \cos(xy) (\frac{d}{dx}(xy)))

Simplifying:

(0 = x \frac{dy}{dx} \sin(xy) + y \sin(xy) + xy \cos(xy) (y + x \frac{dy}{dx}))

Isolating (\frac{dy}{dx}):

(x \frac{dy}{dx} \sin(xy) + xy \cos(xy) (x \frac{dy}{dx}) = - y \sin(xy) - xy \cos(xy) y)

(\frac{dy}{dx} (x \sin(xy) + xy \cos(xy)) = -y \sin(xy) - xy \cos(xy) y)

(\frac{dy}{dx} = \frac{-y \sin(xy) - xy \cos(xy) y}{x \sin(xy) + xy \cos(xy)})

This is the implicit derivative of the given equation with respect to (x).

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