## Calculus (3rd Edition)

$R' = \frac{3R}{5x}$
$\frac{d}{dx}(x^3R^5) = \frac{d}{dx}(1)$ $3x^2R^5 + 5x^3R^4R' = 0$ Now we need to solve for $R'$ $5x^3R^4R' = -3x^2R^5$ $R' = -\frac{3x^2R^5}{5x^3R^4}$ We can simplify this by dividing out the $R$ and $x$ to get: $R' = \frac{3R}{5x}$