## Calculus 8th Edition

$f'(x)= 5.2$
$(h+g)' = h' +g'$ therefore lets choose, $h(x) = 5.2x$ $g(x) = 2.3$ $f(x)=5.2x+2.3=h(x)+g(x)$ $f'(x)=(h(x)+g(x))'=h'(x)+g'(x)$ and $\frac{d}{dx}(c)=0$ therefore $\frac{d}{dx}(2.3)=0$ and $\frac{d}{dx}(x^{n})=nx^{n-1}$ therefore $\frac{d}{dx}(5.2x)=5.2\times x^{0}=5.2\times 1=5.2$ meaning $h'(x)=5.2$ $g'(x)=0$ So, $f'(x)= h'(x)+g'(x) = 5.2 + 0 = 5.2$