If x-1-t1-t and y-2t1-t- then d2ydx2 is equal to

Given : x=1 t1+t and y=2t1+t We can write x+y=1 Differentiating w.r.t. x, we get ⇒ 1+dydx=0 ⇒ nbsp;dydx nbsp; = 1 Differentiating w.r.t. x, we get ∴d^2ydx^ ...

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

Byju's Answer

Standard XII

Mathematics

Composite Function

If x=1-t1+t a...

Question

If $x = \frac{1 - t}{1 + t}$ and $y = \frac{2 t}{1 + t}$ , then $\frac{d^{2} y}{d x^{2}}$ is equal to

\frac{2 t}{{(1 + t)}^{2}}

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

\frac{1}{{(1 + t)}^{4}}

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

\frac{2 t^{2}}{{(1 + t)}^{2}}

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

0

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

Open in App

Solution

The correct option is D $0$
Given : $x = \frac{1 - t}{1 + t}$ and $y = \frac{2 t}{1 + t}$
We can write
$x + y = 1$
Differentiating w.r.t. $x,$ we get
$\Rightarrow 1 + \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = - 1$

Differentiating w.r.t. $x,$ we get
$∴ \frac{d^{2} y}{d x^{2}} = 0$

Suggest Corrections

Similar questions

Q. If

x = cos t

and

y = ln t

; then at

t = \frac{π}{2}

;

(\frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2})

is equal to

Q. If

x = \frac{1 - t}{1 + t} : y = \frac{2 t}{1 + t}

then

\frac{d^{2} y}{d x^{2}} =

Q. If

x = 2 cos 2 t, y = sin 2 t

then -

Q. If

x = a t^{2}

and

y = 2 a t

, then

\frac{d^{2} y}{d x^{2}}

t = \frac{1}{2}

Q. If

x = l o g (1 + t^{2})

and

y = t - t a n^{- 1} t

. Then,

\frac{d y}{d x}

is equal to

Join BYJU'S Learning Program

Explore more

Composite Function

Standard XII Mathematics

Join BYJU'S Learning Program

If x=1−t1+t and y=2t1+t, then d2ydx2 is equal to

The correct option is D 0Given : x=1−t1+t and y=2t1+t We can write x+y=1 Differentiating w.r.t. x, we get ⇒1+dydx=0⇒dydx=−1 Differentiating w.r.t. x, we get ∴d2ydx2=0

If $x = \frac{1 - t}{1 + t}$ and $y = \frac{2 t}{1 + t}$ , then $\frac{d^{2} y}{d x^{2}}$ is equal to

The correct option is D $0$
Given : $x = \frac{1 - t}{1 + t}$ and $y = \frac{2 t}{1 + t}$
We can write
$x + y = 1$
Differentiating w.r.t. $x,$ we get
$\Rightarrow 1 + \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = - 1$

Differentiating w.r.t. $x,$ we get
$∴ \frac{d^{2} y}{d x^{2}} = 0$