Given, f(x+y)=2xf(y)+4y(f(x)
Put y=2 we get,
f(x+2)=2x×3+16f(x)
f′(x+2)=16f′(x)+3×2xln2
Now put x=2 we get,
f′(4)=16f′(2)+12ln2…(i)
Similarly, f(y+2)=4f(y)+3×4y
f′(4)=4f′(y)+3×4yln4
f′(4)=4f′(2)+96ln2⋯(ii)
solving eq. (i) and (ii), we get
f′(2)=7ln2
from equation (i), we get
f′(4)=124ln2
Now ⇒14×f′(2)f′(4)
14×7ln2124ln2=248