Calculate: f(x)=((x+4)/(x-1))^7

Expression: $f\left( x \right)={\left( \frac{ x+4 }{ x-1 } \right)}^{7}$

Take the derivative of both sides

$f '\left( x \right)=\frac{ \mathrm{d} }{ \mathrm{d}x} \left( {\left( \frac{ x+4 }{ x-1 } \right)}^{7} \right)$

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 $g=\frac{ x+4 }{ x-1 }$, to find the derivative

$f '\left( x \right)=\frac{ \mathrm{d} }{ \mathrm{d}g} \left( {g}^{7} \right) \times \frac{ \mathrm{d} }{ \mathrm{d}x} \left( \frac{ x+4 }{ x-1 } \right)$

Find the derivative

$f '\left( x \right)=7{g}^{6} \times \frac{ \mathrm{d} }{ \mathrm{d}x} \left( \frac{ x+4 }{ x-1 } \right)$

Find the derivative of the quotient

$f '\left( x \right)=7{g}^{6} \times \frac{ \left( x-1 \right)-\left( x+4 \right) }{ {\left( x-1 \right)}^{2} }$

Substitute back $g=\frac{ x+4 }{ x-1 }$

$f '\left( x \right)=7 \times {\left( \frac{ x+4 }{ x-1 } \right)}^{6} \times \frac{ \left( x-1 \right)-\left( x+4 \right) }{ {\left( x-1 \right)}^{2} }$

Simplify the expression

$f '\left( x \right)=-\frac{ 35{\left( x+4 \right)}^{6} }{ {\left( x-1 \right)}^{8} }$