If y-sin-1-1-x-1-x-2- then dy-dx-

Y=sin^ 1√(1+x)+√(1 x)/2 Put x=cosθ ∴ y=sin^ 1√(1+cosθ)+√(1 cosθ)/2 =sin^ 1√(2cos^2(θ/2 ))+√(2sin^2(θ/2 ))/2 =sin^ 1√(2)(cosθ/2+sinθ/2 )/2 =sin^ 1(1/√(2)cosθ ...

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

Byju's Answer

Standard XIII

Mathematics

Differentiation Using Substitution

If y=sin-1√1+...

Question

If $y = s i n^{- 1} \frac{\sqrt{(1 + x)} + \sqrt{(1 - x)}}{2}$ , then $\frac{d y}{d x} =$

\frac{1}{\sqrt{(1 - x^{2})}}

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

- \frac{1}{\sqrt{(1 - x^{2})}}

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

- \frac{1}{2 \sqrt{(1 - x^{2})}}

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

None of these

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

Open in App

Solution

The correct option is C $- \frac{1}{2 \sqrt{(1 - x^{2})}}$
$y = s i n^{- 1} \frac{\sqrt{1 + x} + \sqrt{1 - x}}{2} P u t x = c o s θ ∴ y = s i n^{- 1} \frac{\sqrt{1 + c o s θ} + \sqrt{1 - c o s θ}}{2} = s i n^{- 1} \frac{\sqrt{2 c o s^{2} (\frac{θ}{2})} + \sqrt{2 s i n^{2} (\frac{θ}{2})}}{2} = s i n^{- 1} \frac{\sqrt{2} (c o s \frac{θ}{2} + s i n \frac{θ}{2})}{2} = s i n^{- 1} (\frac{1}{\sqrt{2}} c o s \frac{θ}{2} + \frac{1}{\sqrt{2}} s i n \frac{θ}{2}) = s i n^{- 1} (s i n \frac{π}{4} c o s \frac{θ}{2} + c o s \frac{π}{4} s i n \frac{θ}{2}) = s i n^{- 1} s i n (\frac{π}{4} + \frac{θ}{2}) = \frac{π}{4} + \frac{θ}{2} ∴ y = \frac{π}{4} + \frac{1}{2} c o s^{- 1} x ∴ \frac{d y}{d x} = \frac{1}{2} \frac{- 1}{\sqrt{1 - x^{2}}} = \frac{- 1}{2 \sqrt{1 - x^{2}}}$

Suggest Corrections

Similar questions

Q. If

y = s i n^{- 1} \frac{\sqrt{(1 + x)} + \sqrt{(1 - x)}}{2}

, then

\frac{d y}{d x} =

Q. if

y = ({sin}^{- 1} x)^{2} + k {sin}^{- 1} x

, then

(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x}

Q. If

y = sin (2 {sin}^{- 1} x), t h e n \frac{d y}{d x} =

Q. If

y = x \sqrt{1 - x^{2}} + s i n^{- 1} x

, then

\frac{d y}{d x}

Q. If

y = {sec}^{- 1} (\frac{\sqrt{x} + 1}{\sqrt{x} - 1}) + {sin}^{- 1} (\frac{\sqrt{x} - 1}{\sqrt{x} + 1})

, then

\frac{d y}{d x} =

Join BYJU'S Learning Program

If y=sin−1√(1+x)+√(1−x)2, then dydx=

If $y = s i n^{- 1} \frac{\sqrt{(1 + x)} + \sqrt{(1 - x)}}{2}$ , then $\frac{d y}{d x} =$