Find $\frac{dy}{dx}$ if $y=\sqrt[3]{x^3-5x^{4}7x}.$

$\frac{dy}{dx}=0$

$\frac{dy}{dx}=\frac{1}{7}(3x^2-20x^{3}7)(x^3-5x^{4}7x{)}^{(-\frac{6}{7})}$

$\frac{dy}{dx}=\frac{1}{7}(3x^2-20x^{3}7{)}^{(-\frac{6}{7})}$

$,\frac{dy}{dx}=\frac{1}{7}(x^3-5x^{t}7x{)}^{(-\frac{1}{7})}$