We can solve this problem by using logarithmic differentiation.
ln ( f ( x ) ) = ln ( ( x 2 + 1 ) 7 ( x 4 + 2 ) 5 ( x 6 + 3 ) 3 ( x 8 + 4 ) ) = 7 ln ( x 2 + 1 ) + 5 ln ( x 4 + 2 ) + 3 ln ( x 6 + 3 ) + ln ( x 8 + 4 ) {\displaystyle {\begin{aligned}\ln(f(x))&=\ln \left((x^{2}+1)^{7}(x^{4}+2)^{5}(x^{6}+3)^{3}(x^{8}+4)\right)\\&=7\ln(x^{2}+1)+5\ln(x^{4}+2)+3\ln(x^{6}+3)+\ln(x^{8}+4)\end{aligned}}}
Differentiating implicitly yields
1 f ( x ) f ′ ( x ) = 7 x 2 + 1 ⋅ 2 x + 5 x 4 + 2 ⋅ 4 x 3 + 3 x 6 + 3 ⋅ 6 x 5 + 1 x 8 + 4 ⋅ 8 x 7 {\displaystyle {\begin{aligned}{\frac {1}{f(x)}}f'(x)&={\frac {7}{x^{2}+1}}\cdot 2x+{\frac {5}{x^{4}+2}}\cdot 4x^{3}+{\frac {3}{x^{6}+3}}\cdot 6x^{5}+{\frac {1}{x^{8}+4}}\cdot 8x^{7}\end{aligned}}}
So
f ′ ( x ) = ( x 2 + 1 ) 7 ( x 4 + 2 ) 5 ( x 6 + 3 ) 3 ( x 8 + 4 ) ( 14 x x 2 + 1 + 20 x 3 x 4 + 2 + 18 x 5 x 6 + 3 + 8 x 7 x 8 + 4 ) {\displaystyle {\begin{aligned}f'(x)&=(x^{2}+1)^{7}(x^{4}+2)^{5}(x^{6}+3)^{3}(x^{8}+4)\left({\frac {14x}{x^{2}+1}}+{\frac {20x^{3}}{x^{4}+2}}+{\frac {18x^{5}}{x^{6}+3}}+{\frac {8x^{7}}{x^{8}+4}}\right)\end{aligned}}}
