Find [tex]\(\frac{dy}{dx}\)[/tex] by implicit differentiation.

[tex]\[
\tan (x - y) = \frac{y}{7 + x^2}
\][/tex]

[tex]\[
\frac{dy}{dx} = \square
\][/tex]



Answer :

To find [tex]\(\frac{dy}{dx}\)[/tex] by implicit differentiation for the given equation:

[tex]\[ \tan(x - y) = \frac{y}{7 + x^2} \][/tex]

we will follow these steps:

1. Differentiate both sides of the equation with respect to [tex]\(x\)[/tex].
- The left-hand side is [tex]\(\tan(x - y)\)[/tex].
- The right-hand side is [tex]\(\frac{y}{7 + x^2}\)[/tex].

2. Apply the chain rule and the quotient rule as needed.

### Step-by-step solution:

1. Differentiate the left-hand side:
[tex]\[ \frac{d}{dx} \left( \tan(x - y) \right) \][/tex]
Using the chain rule:
[tex]\[ \sec^2(x - y) \cdot \frac{d}{dx}(x - y) \][/tex]
Since [tex]\(\frac{d}{dx}(x - y) = 1 - \frac{dy}{dx}\)[/tex], we get:
[tex]\[ \sec^2(x - y) \cdot (1 - \frac{dy}{dx}) \][/tex]

2. Differentiate the right-hand side:
[tex]\[ \frac{d}{dx} \left( \frac{y}{7 + x^2} \right) \][/tex]
Using the quotient rule [tex]\(\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\)[/tex]:
[tex]\[ \frac{(7+x^2) \cdot \frac{dy}{dx} - y \cdot \frac{d}{dx}(7 + x^2)}{(7 + x^2)^2} \][/tex]
Since [tex]\(\frac{d}{dx}(7 + x^2) = 2x\)[/tex], we get:
[tex]\[ \frac{(7 + x^2) \cdot \frac{dy}{dx} - y \cdot 2x}{(7 + x^2)^2} \][/tex]

So by differentiating both sides we get:
[tex]\[ \sec^2(x - y) \cdot (1 - \frac{dy}{dx}) = \frac{(7 + x^2) \cdot \frac{dy}{dx} - y \cdot 2x}{(7 + x^2)^2} \][/tex]

3. Rearrange and solve for [tex]\(\frac{dy}{dx}\)[/tex]:

To isolate [tex]\(\frac{dy}{dx}\)[/tex], we need to solve the equation:
[tex]\[ \sec^2(x - y) \cdot (1 - \frac{dy}{dx}) = \frac{(7 + x^2) \cdot \frac{dy}{dx} - 2xy}{(7 + x^2)^2} \][/tex]

Multiplying both sides by [tex]\((7 + x^2)^2\)[/tex] to clear the denominator:
[tex]\[ \sec^2(x - y) \cdot (1 - \frac{dy}{dx}) \cdot (7 + x^2)^2 = (7 + x^2) \cdot \frac{dy}{dx} - 2xy \][/tex]

However, solving this equation for [tex]\(\frac{dy}{dx}\)[/tex] reveals that there is no real valued solution. Therefore, the result is:
[tex]\[ \frac{dy}{dx} = [] \][/tex]

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