## Calculus (3rd Edition)

$\frac{d}{dx}(f(2x+g(x)))=f'(2x+g(x))\times(2x+g(x))'$ $=f'(2x+g(x))\times(2+g'(x))$ When $x=1$, $(f(2x+g(x)))'=$ $f'(2\times1+g(1))\times(2+g'(1))$ $=f'(2+4)\times(2+5)=4\times7=28$