Calculate: f(x)=(6)/(sqrt(4-3x))

Expression: $f\left( x \right)=\frac{ 6 }{ \sqrt{ 4-3x } }$

Take the derivative of both sides

$f '\left( x \right)=\frac{ \mathrm{d} }{ \mathrm{d}x} \left( \frac{ 6 }{ \sqrt{ 4-3x } } \right)$

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

$f '\left( x \right)=\frac{ 9 }{ \sqrt{ 4-3x }\left( 4-3x \right) }$