Question

Differentiate the function w.r.t.x:
$\rm e^{x\log a} + e^{a\log x} + e^{a\log a}$

Answer 1

HOPE IT HELPS U ✌️✌️

Answer 2

We know that from the properties of log and exponential

${e}^{x log(a) } = {e}^{ log( {a}^{x} ) } = {a}^{x} \\ \\ {e}^{a log(x) } = {e}^{ log( {x}^{a} ) } = {x}^{a} \\ \\ {e}^{a log(a) } = {e}^{ log( {a}^{a} ) } = {a}^{a}$
So

$\frac{d}{dx} (\rm e^{x\log a} + e^{a\log x} + e^{a\log a}) \\ \\ = \frac{d}{dx} ( {a}^{x} + {x}^{a} + {a}^{a} ) \\ \\ = > \frac{d( {a}^{x}) }{dx} = {a}^{x} log(a) \\ \\ = > \frac{d( {x}^{a}) }{dx} = a {x}^{a - 1} \\ \\ = > \frac{d( {a}^{a}) }{dx} = 0$
So
$\frac{d}{dx} (\rm e^{x\log a} + e^{a\log x} + e^{a\log a}) \\\\= {a}^{x} log(a) + a {x}^{a - 1}$
Hope it helps you.