Answer
$R' = \frac{3R}{5x}$
Work Step by Step
$\frac{d}{dx}(x^3R^5) = \frac{d}{dx}(1)$
$3x^2R^5 + 5x^3R^4R' = 0$
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:
$R' = \frac{3R}{5x}$
