CameraIcon

SearchIcon

MyQuestionIcon

1

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

Sum of n Terms

If y= tan -...

Question

If $y = {({tan}^{- 1} x)}^{2}$
S.T ${(x^{2} + 1)}^{2} y_{2} + 2 x (x^{2} + 1) y_{1} = 2$

Open in App

Solution

$y = {({tan}^{- 1} x)}^{2}$
Differentitating w.r.t $x$
$\frac{d y}{d x} = 2 {tan}^{- 1} x . \frac{1}{1 + x^{2}}$
$y_{1} = \frac{2 {tan}^{- 1} x}{1 + x^{2}}$
$(1 + x^{2}) y_{1} = 2 {tan}^{- 1} x$
D w.r.t $x$
$(1 + x^{2}) y_{2} + 2 x y_{1} = \frac{2}{1 + x^{2}}$ $X$ by $1 + x^{2}$
$(1 + x^{2}) y_{2} + 2 x (1 + x^{2}) y_{1} = 2$

Suggest Corrections

thumbs-up

0

Similar questions

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

{(x^{2} + 1)}^{2} y_{2} + 2 x (x^{2} + 1) y_{1} = 2

y = (t a n^{- 1} x)^{2}

{(x^{2} + 1)}^{2} y_{2} + 2 x (x^{2} + 1) y_{1} = 2

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

(x^{2} + 1)^{2} y_{2} + 2 x (x^{2} + 1) y_{1}

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

(x^{2} + 1)^{2} y_{2} + 2 x (x^{2} + 1) y_{1} = 2

If $y = (t a n^{- 1} x)^{2}, t h e n (x^{2} + 1)^{2} y_{2} + 2 x (x^{2} + 1) y_{1}$ is equal to

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Geometric Progression

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon