Ify -tan -12x-1-x2-1-1- then d y-d x at x-0 is

The correct option is D none of thesey=tan^ 1(2^x+1 2^x/1+2^x.2^x+1 ) =tan^ 12^(x+1) tan^ 12^x ⇒ y'=x^x+1ln2/1+(2^x+1)^2 2^xln2/1+(2^x)^2⇒ y'(0)= 1/10ln2 ...

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

Byju's Answer

Standard XII

Mathematics

Improper Integrals

Ify =tan -12x...

Question

$I f y = t a n^{- 1} (\frac{2^{x}}{1 + x^{2 x + 1}}), t h e n \frac{d y}{d x} a t x = 0 i s$

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

in 2

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

none of these

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

Open in App

Solution

The correct option is D none of these
$y = t a n^{- 1} (\frac{2^{x + 1} - 2^{x}}{1 + 2^{x} {.2}^{x + 1}}) = t a n^{- 1} 2^{(} x + 1) - t a n^{- 1} 2^{x} \Rightarrow y^{'} = \frac{x^{x + 1} l n 2}{1 + (2^{x + 1})^{2}} - \frac{2^{x} l n 2}{1 + (2^{x})^{2}} \Rightarrow y^{'} (0) = - \frac{1}{10} l n 2$

Suggest Corrections

Similar questions

Q. If

y = {tan}^{- 1} (\frac{2 x - 1}{1 + x - x^{2}})

, then

\frac{d y}{d x}

x = 1

is equal to

Q. If

y = {tan}^{- 1} (\frac{2^{x^{'}}}{1 + 2^{2 x + 1}})

then

\frac{d y}{d x}

x = 0

is?

Q. If

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

, then find

\frac{d y}{d x}

Q. If

y = {tan}^{- 1} [\frac{\sqrt{1 + x^{2}} - 1}{x}]

, then

\frac{d y}{d x}

Q. If

y = {sin}^{- 1} 2 x \sqrt{1 - x^{2}} + {sec}^{- 1} \frac{1}{\sqrt{(1 - x^{2})}},

then

\frac{d y}{d x} .

x = 0

Join BYJU'S Learning Program

Ify=tan−1(2x1+x2x+1),then dydx at x=0 is

The correct option is D none of thesey=tan−1(2x+1−2x1+2x.2x+1)=tan−12(x+1)−tan−12x⇒y′=xx+1ln21+(2x+1)2−2xln21+(2x)2⇒y′(0)=−110ln2

$I f y = t a n^{- 1} (\frac{2^{x}}{1 + x^{2 x + 1}}), t h e n \frac{d y}{d x} a t x = 0 i s$