Question

Question no. 01(v) with solution

Answer 1

e raise to x log x
=x^x
let
y=x^x
taking log both sides we get
log(y) =x*logx
1/y*dy/dx =x*1/x+logx * (1)
dy/dx =y(1+logx)
=x^x(1+logx)

Answer 2

(v) e^(xlogx)

$= \ \textgreater \ \frac{d}{dx} [ e^{xlogx} ]$

$= \ \textgreater \ e^{xlogx} * \frac{d}{dx} [x log x]$

$= \ \textgreater \ ( \frac{d}{dx}[x]. logx + x * \frac{d}{dx} [logx])* e^{xlogx}$

$= \ \textgreater \ e^{xlogx} (1 * logx + \frac{1}{x} * x)$

$= \ \textgreater \ e^{xlogx}(log(x) + 1)$

$= \ \textgreater \ x^{x} (log x + 1)$



Hope this helps!