Answer
(a)$\frac{df}{dx}= 4 x^{ -\frac{3}{4}}$
$\int \frac{df}{dx}dx= 16{x^{\frac{1}{4}}}=16\sqrt[4] x+C$
Work Step by Step
(a)$f(x)= 16 \sqrt[4] x$
$\frac{df}{dx}=\frac{d(16 \sqrt[4] x)}{dx}$
$\frac{df}{dx}=16\frac{d( \sqrt[4] x)}{dx}$
$\frac{df}{dx}=16\frac{d(x^{\frac{1}{4}})}{dx}$
$\frac{df}{dx}= 16(\frac{1}{4}) x^{ \frac{1}{4}-1}$
$\frac{df}{dx}= 4 x^{ -\frac{3}{4}}$
(b)
$\int \frac{df}{dx}dx= \int4x^{-\frac{3}{4}}dx$
$\int \frac{df}{dx}dx= 4 \int x^{-\frac{3}{4}}dx$
$\int \frac{df}{dx}dx= 4 ( \frac{x^{-\frac{3}{4}+1}}{-\frac{3}{4}+1})+C$
$\int \frac{df}{dx}dx= 4 ( \frac{x^{ \frac{1}{}}}{\frac{1}{4}})+C$
$\int \frac{df}{dx}dx= 16{x^{\frac{1}{4}}}=16\sqrt[4] x+C$
