Answer:
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Step-by-step explanation:
To find the derivative \( y' \) implicitly from the equation \( 3x + 5y = x^3 + y \), we'll differentiate both sides of the equation with respect to \( x \) and then solve for \( y' \).
Given:
\[ 3x + 5y = x^3 + y \]
Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}(3x) + \frac{d}{dx}(5y) = \frac{d}{dx}(x^3) + \frac{d}{dx}(y) \]
\[ 3 + 5\frac{dy}{dx} = 3x^2 + \frac{dy}{dx} \]
Now, let's isolate \( \frac{dy}{dx} \) terms:
\[ 5\frac{dy}{dx} - \frac{dy}{dx} = 3x^2 - 3 \]
\[ (5 - 1) \frac{dy}{dx} = 3x^2 - 3 \]
\[ 4 \frac{dy}{dx} = 3x^2 - 3 \]
\[ \frac{dy}{dx} = \frac{3x^2 - 3}{4} \]
So, \( y' = \frac{3x^2 - 3}{4} \) is the derivative of \( y \) implicitly from the given equation.
