Question

If $f\left(x\right)=\frac{a^x}{x^a}$ then $f^{\prime }\left(a\right)=$

• A. $\log a-1$
• B. $\log a-a$
• C. $a\log a-a$
• D. $aloga+a$

Answer 1

The correct option is A $\log a-1$

$f\left(x\right)=\left(\frac{a^x}{x^a}\right)f^{\prime }\left(x\right)=\left(\frac{a^{x}loga.x^a-a^x\times a{x}^{a-1}}{(x^a{)}^2}\right)$

$\therefore f^{\prime }\left(a\right)=\left(\frac{a^{a}loga.a^a-a^a\times a{a}^{a-1}}{(a^a{)}^2}\right)$

$=\left(\frac{\left(a^{2a}\right)(loga-1)}{\left(a^{2a}\right)}\right)=(loga-1)$