The correct option is A 1√(2ax x^2) ddx[tan^ 1√(2ax x^2)a x] (a x)^2(a x)^2+2ax x^2×(a x)(2a 2x)2(2ax x^2)^32+√(2ax x^2)(a x)^2 (a x)^2a^2× ...

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

Derivative of Standard Inverse Trigonometric Functions

ddx[tan-1√2ax...

Question

$\frac{d}{d x} [{tan}^{- 1} \frac{\sqrt{2 a x - x^{2}}}{a - x}] = ?$

\frac{1}{\sqrt{a^{2} - x^{2}}}

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

\frac{1}{\sqrt{2 a x - x^{2}}}

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

\frac{- 1}{\sqrt{2 a x - x^{2}}}

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

\frac{- 1}{\sqrt{a^{2} - x^{2}}}

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

Open in App

Solution

The correct option is A $\frac{1}{\sqrt{2 a x - x^{2}}}$
$\frac{d}{d x} [{tan}^{- 1} \frac{\sqrt{2 a x - x^{2}}}{a - x}]$
$\frac{(a - x)^{2}}{(a - x)^{2} + 2 a x - x^{2}} \times \frac{\frac{(a - x) (2 a - 2 x)}{2 (2 a x - x^{2})^{\frac{3}{2}}} + \sqrt{2 a x - x^{2}}}{(a - x)^{2}}$
$\frac{(a - x)^{2}}{a^{2}} \times \frac{a^{2}}{(a - x)^{2} \sqrt{2 a x - x^{2}}}$
$\frac{1}{\sqrt{2 a x - x^{2}}}$

Suggest Corrections

Similar questions

x^{2} + 2 a x + (a^{2} - 1) = 0

\int (\frac{(x^{2} + 2) a^{(x + {tan}^{- 1} x)}}{x^{2} + 1}) d x =

x^{2} - 1 + 2 a x + a^{2}

a^{2} - 2 a x + x^{2} + 2 a - 2 x + 1

If $x^{2} - 1 = - b^{2} - 2 b x a n d x^{2} - 1 = - a^{2} - 2 a x$ have exactly one root in common then

Join BYJU'S Learning Program