Question

please solve with solution ​

Answer 1

Step-by-step explanation:

We have,

$f(x) = \frac{ {a}^{x} }{ {x}^{a} }$

Now, differentiating both sides w.r.t. x,

$\frac{d \{f(x) \}}{dx} = \frac{d}{dx} \bigg( \frac{ {a}^{x} }{ {x}^{a} } \bigg)$

$\implies \: f ^{ \prime}(x) = \frac{ {x}^{a} .\frac{d}{dx}( {a}^{x}) - {a}^{x} . \frac{d}{dx}( {x}^{a} ) }{ ({x}^{a} )^{2} }$

$\implies \: f ^{ \prime}(x) = \frac{ {x}^{a} . {a}^{x}. ln(a) - {a}^{x} . a. {x}^{a - 1} }{ {x}^{2a} }$

Now,

$\implies \: f ^{ \prime}(a) = \frac{ {a}^{a} . {a}^{a}. ln(a) - {a}^{a} . a. {a}^{a - 1} }{ {a}^{2a} }$

$\implies \: f ^{ \prime}(a) = \frac{ {a}^{a + a} . ln(a) - {a}^{a + 1 + a - 1} }{ {a}^{2a} }$

$\implies \: f ^{ \prime}(a) = \frac{ {a}^{2a } . ln(a) - {a}^{2a} }{ {a}^{2a} }$

$\implies \: f ^{ \prime}(a) = \frac{ {a}^{2a } . \{ln(a) - 1\} }{ {a}^{2a} }$

$\implies \: f ^{ \prime}(a) = ln(a) - 1$

So,

$\rm \large\:\bold{f^{\prime}(a)=ln(a)-1}$