# Solve for: f(x)=(3+ln(x))^8

## Expression: $f\left( x \right)={\left( 3+\ln\left({x}\right) \right)}^{8}$

Take the derivative of both sides

$f '\left( x \right)=\frac{ \mathrm{d} }{ \mathrm{d}x} \left( {\left( 3+\ln\left({x}\right) \right)}^{8} \right)$

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

$f '\left( x \right)=\frac{ 8{\left( 3+\ln\left({x}\right) \right)}^{7} }{ x }$

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