Answer
$$f'\left( x \right) = 5{\left( {5x + 29} \right)^4}\left( {90x + 127} \right)$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {\left( {5x + 29} \right)^5}\left( {15x + 8} \right) \cr
& {\text{Calculate the derivative of the function}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left( {{{\left( {5x + 29} \right)}^5}\left( {15x + 8} \right)} \right) \cr
& {\text{use the product rule }} \cr
& f'\left( x \right) = \left( {15x + 8} \right)\frac{d}{{dx}}\left( {{{\left( {5x + 29} \right)}^5}} \right) + {\left( {5x + 29} \right)^5}\frac{d}{{dx}}\left( {\left( {15x + 8} \right)} \right) \cr
& {\text{use the chain rule to }}\frac{d}{{dx}}\left( {{{\left( {5x + 29} \right)}^5}} \right) \cr
& f'\left( x \right) = \left( {15x + 8} \right)\left( 5 \right){\left( {5x + 29} \right)^4}\frac{d}{{dx}}\left( {5x + 29} \right) + {\left( {5x + 29} \right)^5}\frac{d}{{dx}}\left( {\left( {15x + 8} \right)} \right) \cr
& {\text{compute derivatives}} \cr
& f'\left( x \right) = 5\left( {15x + 8} \right){\left( {5x + 29} \right)^4}\left( 5 \right) + {\left( {5x + 29} \right)^5}\left( {15} \right) \cr
& f'\left( x \right) = 25\left( {15x + 8} \right){\left( {5x + 29} \right)^4} + 15{\left( {5x + 29} \right)^5} \cr
& {\text{factor}} \cr
& f'\left( x \right) = 5{\left( {5x + 29} \right)^4}\left( {5\left( {15x + 8} \right) + 3\left( {5x + 29} \right)} \right) \cr
& f'\left( x \right) = 5{\left( {5x + 29} \right)^4}\left( {75x + 40 + 15x + 87} \right) \cr
& f'\left( x \right) = 5{\left( {5x + 29} \right)^4}\left( {90x + 127} \right) \cr} $$
