The value of (d/dx) [xn loga xex ] =

Question

The value of (d/dx) [xn loga xex ] =  (A) ex loga x + (xn-1/ loge a) (B)  ex xn-1 [x loga x + (1/ logea) + x logax ] (C) nxn-1 loga (xex)  (D) xn loga (xex)

Tardigrade Admin · Accepted Answer

(d/dx) [xn ( loga x)ex] =xn loga x. ex +xnex. (1/x) logae + nxn-1 logax . ex = ex [xn loga x+xn-1 loga e + nxn-1 logax] = x n-1 ex [x loga x + (1/ logea) + n logax]

Q. The value of $\frac{d}{d x} [x^{n} lo g_{a} x e^{x}] =$

$e^{x} lo g_{a} x + \frac{x ^{n - 1}}{l o g _{e} a}$

$e^{x} x^{n - 1} [x lo g_{a} x + \frac{1}{l o g _{e} a} + x lo g_{a} x]$

$n x^{n - 1} lo g_{a} (x e^{x})$

$x^{n} lo g_{a} (x e^{x})$

$\frac{d}{d x} [x^{n} (lo g_{a} x) e^{x}]$
$= x^{n} lo g_{a} x . e^{x} + x^{n} e^{x} . \frac{1}{x} lo g_{a} e + n x^{n - 1} lo g_{a} x . e^{x}$
$= e^{x} [x^{n} lo g_{a} x + x^{n - 1} lo g_{a} e + n x^{n - 1} lo g_{a} x]$
$= x^{n - 1} e^{x} [x lo g_{a} x + \frac{1}{l o g _{e} a} + n lo g_{a} x]$

Q. The value of dxd​[xnloga​xex]=

exloga​x+loge​axn−1​

exxn−1[xloga​x+loge​a1​+xloga​x]

nxn−1loga​(xex)

xnloga​(xex)