Answer
$f^{'}(x)=(-2)(5.7x^2+3.5x+2.9)^3 (3.8x^2+5.2x+7)^{-3} (7.6x+5.2)+3(3.8x^2+5.2x+7)^{-2}(5.7x^2+3.5x+2.9)^{2} (11.4x+3.5)$
Work Step by Step
$f(x)=(5.7x^2+3.5x+2.9)^3 (3.8x^2+5.2x+7)^{-2}$
$f^{'}(x)=(5.7x^2+3.5x+2.9)^3 \frac{ d(3.8x^2+5.2x+7)^{-2} }{dx} +(3.8x^2+5.2x+7)^{-2}\frac{d(5.7x^2+3.5x+2.9)^3}{dx} $
$f^{'}(x)=(5.7x^2+3.5x+2.9)^3 (-2) (3.8x^2+5.2x+7)^{-2-1} \frac{d(3.8x^2+5.2x+7) }{dx} +(3.8x^2+5.2x+7)^{-2}(3)(5.7x^2+3.5x+2.9)^{3-1} \frac{d( 5.7x^2+3.5x+2.9 )}{dx}$
$f^{'}(x)=(5.7x^2+3.5x+2.9)^3 (-2) (3.8x^2+5.2x+7)^{-3} (2\times3.8x+5.2)+(3.8x^2+5.2x+7)^{-2}(3)(5.7x^2+3.5x+2.9)^{2} (2\times5.7x+3.5)$
$f^{'}(x)=(-2)(5.7x^2+3.5x+2.9)^3 (3.8x^2+5.2x+7)^{-3} (7.6x+5.2)+3(3.8x^2+5.2x+7)^{-2}(5.7x^2+3.5x+2.9)^{2} (11.4x+3.5)$
