Answer:
if you don't understand please find the attachment
Step-by-step explanation:
We have, y=x ^x
Taking log on both the sides, we get
logy=xlogx
On differentiating w.r.t. x, we get
=
x
x
+logx
⇒
dx
dy
=y+ylogx
⇒
dx
dy
=x x
(1+logx) ....(∵y=x ^x )
