Question

If $y=e^{x\log a}+e^{a\log x}+e^{a\log a}$, then $\frac{dy}{dx}=$

• A. $a^x\log a+x^{a-1}$
• B. $a^x\log a+ax$
• C. $a^x\log a+a{x}^{a-1}$
• D. $a^x\log a+a{x}^a$

Answer 1

The correct option is C $a^x\log a+a{x}^{a-1}$

Let $y=e^{xloga}+e^{alogx}+e^{aloga}$
$=e^{{\log }_e{a}^x}+e^{{\log }_e{x}^a}+e^{{\log }_e{a}^a}$
$=a^x+x^a+a^a$ ......... Since $e^{{\log }_{e}x}=x$
$\frac{dy}{dx}=\frac{d}{dx}(a^x+x^a+a^a)$
$=a^x\log a+a.x^{a-1}$