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]
