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

Question

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

Christopher Allers · Accepted Answer

#(dy)/(dx)=-y/x#

We canwrite #1=xysin(xy)# as

#xy=csc(xy)#

Hence implict derivative is given by

#y+x(dy)/(dx)=-csc(xy)cot(xy)(y+x(dy)/(dx))#

or #(dy)/(dx)(x(1+csc(xy)cot(xy))=-y(1+csc(xy)cot(xy))#

or #(dy)/(dx)=-y/x#

Charles Arocha · Answer

undefined 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 $x$.