Answer
$f^{'}(x)=(12.6)[4.8^x \ln 4.8]x^{-2}-25.2(4.8^x)(x^{-3})$
Work Step by Step
$f(x)=\frac{12.6(4,8^x)}{x^2}$
$f(x)=12.6(4.8^x)x^{-2}$
Taking derivative with respect to x, using product rule
$f^{'}(x)=[\frac{d(12.6)(4.8^x)}{dx}]x^{-2}+12.6(4.8^x)\frac{d(x^{-2})}{dx}$
$f^{'}(x)=(12.6)[\frac{d(4.8^x)}{dx}]x^{-2}+12.6(4.8^x)(-2x^{-3})$
$f^{'}(x)=(12.6)[4.8^x \ln 4.8]x^{-2}+12.6(4.8^x)(-2x^{-3})$
$f^{'}(x)=(12.6)[4.8^x \ln 4.8]x^{-2}-25.2(4.8^x)(x^{-3})$
