Differentiate $$ \frac{y-x}{1+x y} $$

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UpStudy AI · Accepted Answer

It seems that the expression $$ \frac{y-x}{1+xy} $$ is not directly supported for differentiation. However, we can differentiate it using the quotient rule manually.

The quotient rule states that if you have a function $$ \frac{u}{v} $$, the derivative is given by:

$$
\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}
$$

where $$ u = y - x $$ and $$ v = 1 + xy $$.

### Step 1: Identify $$ u $$ and $$ v $$

- $$ u = y - x $$
- $$ v = 1 + xy $$

### Step 2: Differentiate $$ u $$ and $$ v $$

1. **Differentiate $$ u $$**:
   $$
   u' = \frac{d}{dx}(y - x) = 0 - 1 = -1
   $$

2. **Differentiate $$ v $$**:
   $$
   v' = \frac{d}{dx}(1 + xy) = 0 + y = y
   $$

### Step 3: Apply the Quotient Rule

Now, we can apply the quotient rule:

$$
\frac{d}{dx}\left(\frac{y-x}{1+xy}\right) = \frac{(-1)(1+xy) - (y-x)(y)}{(1+xy)^2}
$$

### Step 4: Simplify the Expression

Now, let's simplify the numerator:

$$
= \frac{-(1 + xy) - (y^2 - xy)}{(1 + xy)^2}
$$

Combine like terms:

$$
= \frac{-1 - xy - y^2 + xy}{(1 + xy)^2}
$$

This simplifies to:

$$
= \frac{-1 - y^2}{(1 + xy)^2}
$$

### Final Result

Thus, the derivative of $$ \frac{y-x}{1+xy} $$ with respect to $$ x $$ is:

$$
\frac{d}{dx}\left(\frac{y-x}{1+xy}\right) = \frac{-1 - y^2}{(1 + xy)^2}
$$
The derivative of $$ \frac{y - x}{1 + xy} $$ with respect to $$ x $$ is $$ \frac{-1 - y^2}{(1 + xy)^2} $$.

Differentiate \( \frac{y-x}{1+x y} \)

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