How do you find #(dy)/(dx)# given #xy=tanxy#?

Question

Paisley Johnson · Accepted Answer

#-y/x#

#xy=tanxy->y = y(x)# so

#f(y(x),x) = y(x)x-tan(y(x)x)=0#

now

#d/(dx)f(y(x),x) = y(x) + x y(x) - Sec(x y(x))^2 (y(x) + x y'(x)) = 0#

Solving for #y'(x)# we obtain

#dy/dx = -y/x#

We can also obtain the same result straightforward

#dy/dx = - (f_x)/(f_y) = -(y - y Sec(x y)^2)/(x - x Sec(x y)^2) = -y/x#

Scarlett Allison · Answer

undefined To find $ \frac{dy}{dx} $ given $ xy = \tan(xy) $, you can use implicit differentiation.

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

The steps are as follows:

1. Start with $ xy = \tan(xy) $.
2. Differentiate both sides with respect to $x$:

$$ \frac{d}{dx}(xy) = \frac{d}{dx}(\tan(xy)) $$

3. Apply the product rule and chain rule:

$$ y + x\frac{dy}{dx} = \sec^2(xy)\left(y + x\frac{dy}{dx}\right) $$

4. Solve for $ \frac{dy}{dx} $:

$$ y + x\frac{dy}{dx} = \sec^2(xy)y + x\sec^2(xy)\frac{dy}{dx} $$
   $$ x\frac{dy}{dx} - x\sec^2(xy)\frac{dy}{dx} = \sec^2(xy)y - y $$
   $$ x\frac{dy}{dx}(1 - \sec^2(xy)) = \sec^2(xy)y - y $$
   $$ x\frac{dy}{dx}(-\tan^2(xy)) = \sec^2(xy)y - y $$
   $$ \frac{dy}{dx} = \frac{\sec^2(xy)y - y}{-x\tan^2(xy)} $$

So, $ \frac{dy}{dx} = \frac{\sec^2(xy)y - y}{-x\tan^2(xy)} $.