Answer
(a)$\frac{df}{dx}=28x -1 $
(b)$\int \frac{df}{dx} dx= \int df= f(x)=14{x^2}-x+C$
Work Step by Step
$f(x)=14x^2-x+15$
(a)
$\frac{df}{dx}=\frac{d(14x^2-x+15)}{dx}$
$\frac{df}{dx}=\frac{d(14x^2)}{dx} -\frac{d(x)}{dx}+\frac{d(15)}{dx} $
$\frac{df}{dx}=14\frac{d(x^2)}{dx} -\frac{d(x)}{dx}+\frac{d(15)}{dx} $
$\frac{df}{dx}=14(2x) -1+0 $
$\frac{df}{dx}=28x -1 $
(b)
$\int \frac{df}{dx} dx= \int (28x-1)dx$
$\int \frac{df}{dx} dx= 28\int x dx-\int dx$
$\int \frac{df}{dx} dx= 28 (\frac{x^2}{2})-x+C$
$\int \frac{df}{dx} dx= 14{x^2}-x+C$
$\int \frac{df}{dx} dx= \int df= f(x)=14{x^2}-x+C$
