$f '\left( x \right)=\frac{ \mathrm{d} }{ \mathrm{d}x} \left( \frac{ 6 }{ \sqrt{ 4-3x } } \right)$
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} }$$f '\left( x \right)=-6 \times \frac{ \frac{ \mathrm{d} }{ \mathrm{d}x} \left( \sqrt{ 4-3x } \right) }{ {\sqrt{ 4-3x }}^{2} }$
Use the chain rule $\frac{ \mathrm{d} }{ \mathrm{d}x} \left( f\left( g \right) \right)=\frac{ \mathrm{d} }{ \mathrm{d}g} \left( f\left( g \right) \right) \times \frac{ \mathrm{d} }{ \mathrm{d}x} \left( g \right)$, where $g=4-3x$, to find the derivative$f '\left( x \right)=-6 \times \frac{ \frac{ \mathrm{d} }{ \mathrm{d}g} \left( \sqrt{ g } \right) \times \frac{ \mathrm{d} }{ \mathrm{d}x} \left( 4-3x \right) }{ {\sqrt{ 4-3x }}^{2} }$
Find the derivative$f '\left( x \right)=-6 \times \frac{ \frac{ 1 }{ 2\sqrt{ g } } \times \frac{ \mathrm{d} }{ \mathrm{d}x} \left( 4-3x \right) }{ {\sqrt{ 4-3x }}^{2} }$
Find the derivative of the sum or difference$f '\left( x \right)=-6 \times \frac{ \frac{ 1 }{ 2\sqrt{ g } } \times \left( -3 \right) }{ {\sqrt{ 4-3x }}^{2} }$
Substitute back $g=4-3x$$f '\left( x \right)=-6 \times \frac{ \frac{ 1 }{ 2\sqrt{ 4-3x } } \times \left( -3 \right) }{ {\sqrt{ 4-3x }}^{2} }$
Simplify the expression$f '\left( x \right)=\frac{ 9 }{ \sqrt{ 4-3x }\left( 4-3x \right) }$