Calculate: y=(2)/((12x^4-7x^3)^5)

Expression: $y=\frac{ 2 }{ {\left( 12{x}^{4}-7{x}^{3} \right)}^{5} }$

Take the derivative of both sides

$y '=\frac{ \mathrm{d} }{ \mathrm{d}x} \left( \frac{ 2 }{ {\left( 12{x}^{4}-7{x}^{3} \right)}^{5} } \right)$

Use differentiation rule $\frac{ \mathrm{d} }{ \mathrm{d}x} \left( \frac{ a }{ f } \right)=-a \times \frac{ \frac{ \mathrm{d} }{ \mathrm{d}x} \left( f \right) }{ {f}^{2} }$

$y '=-2 \times \frac{ \frac{ \mathrm{d} }{ \mathrm{d}x} \left( {\left( 12{x}^{4}-7{x}^{3} \right)}^{5} \right) }{ {\left( {\left( 12{x}^{4}-7{x}^{3} \right)}^{5} \right)}^{2} }$

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 $g=12{x}^{4}-7{x}^{3}$, to find the derivative

$y '=-2 \times \frac{ \frac{ \mathrm{d} }{ \mathrm{d}g} \left( {g}^{5} \right) \times \frac{ \mathrm{d} }{ \mathrm{d}x} \left( 12{x}^{4}-7{x}^{3} \right) }{ {\left( {\left( 12{x}^{4}-7{x}^{3} \right)}^{5} \right)}^{2} }$

Find the derivative

$y '=-2 \times \frac{ 5{g}^{4} \times \frac{ \mathrm{d} }{ \mathrm{d}x} \left( 12{x}^{4}-7{x}^{3} \right) }{ {\left( {\left( 12{x}^{4}-7{x}^{3} \right)}^{5} \right)}^{2} }$

Find the derivative of the sum or difference

$y '=-2 \times \frac{ 5{g}^{4} \times \left( 12 \times 4{x}^{3}-7 \times 3{x}^{2} \right) }{ {\left( {\left( 12{x}^{4}-7{x}^{3} \right)}^{5} \right)}^{2} }$

Substitute back $g=12{x}^{4}-7{x}^{3}$

$y '=-2 \times \frac{ 5{\left( 12{x}^{4}-7{x}^{3} \right)}^{4} \times \left( 12 \times 4{x}^{3}-7 \times 3{x}^{2} \right) }{ {\left( {\left( 12{x}^{4}-7{x}^{3} \right)}^{5} \right)}^{2} }$

Simplify the expression

$y '=\frac{ 210{x}^{2}-480{x}^{3} }{ {\left( 12{x}^{4}-7{x}^{3} \right)}^{6} }$