Let u = x.lnx, , w= x and t = lnx; w' =1 ; t' = 1/x
f(x) = e^(x.lnx) ; f(u) = e^(u); f'(u) = u'.e^(u)
let' find the derivative u' of u
u = w.t
u'= w't + t'w; u' = lnx + x/x = lnx+1
u' = x+1 and f'(u) = ln(x+1).e^(xlnx)
finally the derivative of f(x) =ln(x+1).e^(x.lnx) + 2x
I’m not sure what the e answer is
Answer:
Step-by-step explanation:
if (x,y) is the centroid. Then x=(x1+x2+x3)/3,y=(y1+y2+y3)/3
5.
x=(-2+4+10)/3=4
y=(6+0+6)/3=4
centroid=(4,4)
6.
x=(3+5-2)/3=2
y=(3-1+1)/3=1
centroid=(2,1)
