Evaluate: y=\sqrt[3]{(2x^3-7x)^2}

Expression: $y=\sqrt[3]{{\left( 2{x}^{3}-7x \right)}^{2}}$

Take the natural logarithm of both sides of the equation

$\ln\left({y}\right)=\ln\left({\sqrt[3]{{\left( 2{x}^{3}-7x \right)}^{2}}}\right)$

Simplify the expression

$\ln\left({y}\right)=\frac{ 2 }{ 3 } \times \ln\left({2{x}^{3}-7x}\right)$

Differentiate both sides of the equation implicitly with respect to $x$

$\frac{ \mathrm{d} }{ \mathrm{d}x} \left( \ln\left({y}\right) \right)=\frac{ \mathrm{d} }{ \mathrm{d}x} \left( \frac{ 2 }{ 3 } \times \ln\left({2{x}^{3}-7x}\right) \right)$

Use the chain rule $\frac{ \mathrm{d} }{ \mathrm{d}x} \left( \ln\left({y}\right) \right)=\frac{ \mathrm{d} }{ \mathrm{d}y} \left( \ln\left({y}\right) \right) \times \frac{ \mathrm{d}y }{ \mathrm{d}y }$ to find the derivative

$\frac{ \mathrm{d} }{ \mathrm{d}y} \left( \ln\left({y}\right) \right) \times \frac{ \mathrm{d}y }{ \mathrm{d}y }=\frac{ \mathrm{d} }{ \mathrm{d}x} \left( \frac{ 2 }{ 3 } \times \ln\left({2{x}^{3}-7x}\right) \right)$

Use differentiation rule $\frac{ \mathrm{d} }{ \mathrm{d}x} \left( a \times f \right)=a \times \frac{ \mathrm{d} }{ \mathrm{d}x} \left( f \right)$

$\frac{ \mathrm{d} }{ \mathrm{d}y} \left( \ln\left({y}\right) \right) \times \frac{ \mathrm{d}y }{ \mathrm{d}y }=\frac{ 2 }{ 3 } \times \frac{ \mathrm{d} }{ \mathrm{d}x} \left( \ln\left({2{x}^{3}-7x}\right) \right)$

Use $\frac{ \mathrm{d} }{ \mathrm{d}x} \left( \ln\left({x}\right) \right)=\frac{ 1 }{ x }$ to find derivative

$\frac{ 1 }{ y } \times \frac{ \mathrm{d}y }{ \mathrm{d}y }=\frac{ 2 }{ 3 } \times \frac{ \mathrm{d} }{ \mathrm{d}x} \left( \ln\left({2{x}^{3}-7x}\right) \right)$

Use the chain rule $\frac{ \mathrm{d} }{ \mathrm{d}x} \left( f\left( g \right) \right)=\frac{ \mathrm{d} }{ \mathrm{d}g} \left( f\left( g \right) \right) \times \frac{ \mathrm{d} }{ \mathrm{d}x} \left( g \right)$, where $g=2{x}^{3}-7x$, to find the derivative

$\frac{ 1 }{ y } \times \frac{ \mathrm{d}y }{ \mathrm{d}y }=\frac{ 2 }{ 3 } \times \frac{ \mathrm{d} }{ \mathrm{d}g} \left( \ln\left({g}\right) \right) \times \frac{ \mathrm{d} }{ \mathrm{d}x} \left( 2{x}^{3}-7x \right)$

Find the derivative

$\frac{ 1 }{ y } \times \frac{ \mathrm{d}y }{ \mathrm{d}y }=\frac{ 2 }{ 3 } \times \frac{ 1 }{ g } \times \frac{ \mathrm{d} }{ \mathrm{d}x} \left( 2{x}^{3}-7x \right)$

Find the derivative of the sum or difference

$\frac{ 1 }{ y } \times \frac{ \mathrm{d}y }{ \mathrm{d}y }=\frac{ 2 }{ 3 } \times \frac{ 1 }{ g } \times \left( 2 \times 3{x}^{2}-7 \right)$

Substitute back $g=2{x}^{3}-7x$

$\frac{ 1 }{ y } \times \frac{ \mathrm{d}y }{ \mathrm{d}y }=\frac{ 2 }{ 3 } \times \frac{ 1 }{ 2{x}^{3}-7x } \times \left( 2 \times 3{x}^{2}-7 \right)$

Simplify the expression

$\frac{ 1 }{ y } \times \frac{ \mathrm{d}y }{ \mathrm{d}y }=\frac{ 12{x}^{2}-14 }{ 6{x}^{3}-21x }$

Multiply both sides of the equation by $y$

$\frac{ \mathrm{d}y }{ \mathrm{d}y }=y \times \frac{ 12{x}^{2}-14 }{ 6{x}^{3}-21x }$

Substitute the initial equation $y=\sqrt[3]{{\left( 2{x}^{3}-7x \right)}^{2}}$ to express the derivative only in terms of $x$

$\frac{ \mathrm{d}y }{ \mathrm{d}y }=\sqrt[3]{{\left( 2{x}^{3}-7x \right)}^{2}} \times \frac{ 12{x}^{2}-14 }{ 6{x}^{3}-21x }$

Calculate the product

$\frac{ \mathrm{d}y }{ \mathrm{d}y }=\frac{ \sqrt[3]{{\left( 2{x}^{3}-7x \right)}^{2}}\left( 12{x}^{2}-14 \right) }{ 6{x}^{3}-21x }$