$\frac{d}{d x} (\frac{1}{x \sqrt{x}}) =$

$\frac{- 3}{2} x^{- 3 / 2}$

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

$\frac{- 3}{2} x^{- 5 / 2}$

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

$x^{- 5 / 2}$

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

None of the above

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

Open in App

Solution

The correct option is B
$\frac{- 3}{2} x^{- 5 / 2}$

As mentioned in the videos that derivative of a function in the form of $x^{n} = n x^{n - 1} .$
Now looking at this function closely we can say that it is nothing but $\frac{1}{x^{1 + 1 / 2}} = \frac{1}{x^{3 / 2}} = x^{- 3 / 2} .$
So, $\frac{d}{d x} (x^{- 3 / 2}) = \frac{- 3}{2} x^{(- 3 / 2 - 1)}$ (Using the above mentioned formula).
$= \frac{- 3}{2} x^{- 5 / 2}$ , which is the correct answer.

Suggest Corrections

Similar questions

\frac{d}{d x} (\frac{1 + cot x}{1 - cot x}) =

\frac{d}{d x} (\frac{1}{\sqrt{x}})

\frac{d}{d x} \frac{1}{\sqrt{x + a}}

$I f | x | < 1 t h e n \frac{d}{d x} (1 - 2 x + 3 x^{2} - . . . . . . . .) =$

\frac{d}{d x} (\frac{1 + x^{2} + x^{4}}{1 + x + x^{2}}) = a x + b, t h e n (a, b) =

Join BYJU'S Learning Program