The correct option is D log a a According to question, f' (x)= a^x loga x^a x^a+1 a^x #160; x^2a ⇒ #160; #160; f' (a)= a^a loga. a^a ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Logarithmic Differentiation

If fx=axxa ...

Question

If $f (x) = \frac{a^{x}}{x^{a}}$ then $f^{'} (a) =$ ?

l o g a - 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

l o g a - a

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

a l o g a - a

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a l o g a + a

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $l o g a - a$
According to question,
$f^{'} (x) = \frac{a^{x} l o g a x^{a} - x^{a + 1} a^{x}}{x^{2 a}}$
$\Rightarrow f^{'} (a) = \frac{a^{a} l o g a . a^{a} - a^{a + 1} a^{a}}{a^{2 a}}$
Hence,
$\Rightarrow f^{'} (a) = l o g a - a$

Suggest Corrections

Similar questions

lim x \to 0 \frac{a^{x} - 1}{x} = log a

lim x \to 0 \frac{2^{x} - 1}{{(1 + x)}^{1 / 2} - 1}

= a log a

a

(a > 1, x > 1)

(0 < a < 1, 0 < x < 1)

{log}_{a} x > 0

{log}_{a} x

(0 < a < 1, x > 1)

(a > 1, 0 < x < 1)

{log}_{a} x < 0

a > b

{log}_{b} a > 1

{log}_{a} b < 1.

\frac{{log}_{a} (3.162)}{{log}_{a} (2 / 3)} .

log \frac{a}{b} + log \frac{b}{a} = log (a + b),

lim x \to \infty \frac{c o t^{- 1} (x^{- a} {log}_{a} x)}{{sec}^{- 1} (a^{x} {log}_{x} a)}, (a > 1)

\int_{0}^{60} \frac{d x}{2 x + 1} = log a

a =

Join BYJU'S Learning Program