$f '\left( x \right)=\frac{ \mathrm{d} }{ \mathrm{d}x} \left( {\left( 3+\ln\left({x}\right) \right)}^{8} \right)$
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 $g=3+\ln\left({x}\right)$, to find the derivative$f '\left( x \right)=\frac{ \mathrm{d} }{ \mathrm{d}g} \left( {g}^{8} \right) \times \frac{ \mathrm{d} }{ \mathrm{d}x} \left( 3+\ln\left({x}\right) \right)$
Find the derivative$f '\left( x \right)=8{g}^{7} \times \frac{ \mathrm{d} }{ \mathrm{d}x} \left( 3+\ln\left({x}\right) \right)$
Find the derivative of the sum or difference$f '\left( x \right)=8{g}^{7} \times \frac{ 1 }{ x }$
Substitute back $g=3+\ln\left({x}\right)$$f '\left( x \right)=8{\left( 3+\ln\left({x}\right) \right)}^{7} \times \frac{ 1 }{ x }$
Calculate the product$f '\left( x \right)=\frac{ 8{\left( 3+\ln\left({x}\right) \right)}^{7} }{ x }$