If y-sin sin x and d2 y-d x2-d y-d xtan x-fx-0- then fx equals-

The correct option is C dy/dx=cos (sin x)cos x d^2y/dx^2= I I= cos (sin x)sin x cos x [sin(sin x)]cos x I +dy/dxtan x= cos^2 x sin(sin x) d^2y/dx^2+dy/dxtan ...

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

Byju's Answer

Standard XI

Mathematics

L'Hospital Rule to Remove Indeterminate Form

If y=sin sin ...

Question

If y = sin (sin x) and $\frac{d^{2} y}{d x^{2}} + \frac{d y}{d x}$ tan x+f(x) = 0, then f(x) equals.

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

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

Open in App

Solution

The correct option is D
$\frac{d y}{d x} = c o s (s i n x) c o s x$
$\frac{d^{2} y}{d x^{2}}$ = I
I=- $cos (s i n x) s i n x - c o s x [s i n (s i n x)] c o s x$
I + $\frac{d y}{d x} t a n x = - c o s^{2} x s i n (s i n x)$
$\frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} t a n x + c o s^{2} x s i n (s i n x) = 0$
$f (x) = c o s^{2} x s i n (s i n x)$

Suggest Corrections

Similar questions

Q. If

y = s i n (s i n x) a n d \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} t a n x + f (x) = 0, t h e n f (x) e q u a l s

Q. If the solution of

y = x \frac{d y}{d x} + \frac{d y}{d x} - {(\frac{d y}{d x})}^{2}; \frac{d^{2} y}{d x^{2}} \neq 0

y = f (x)

and it is defined as

f : R \to R

then

f (x)

Q.
lf

y = sin (sin x)

and

\frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} tan x + f (x) = 0

, then

f (x)

equals to

Q. If

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

Q. If

y = f (x), t h e n \frac{d^{2} y}{d x^{2}} = (f^{'} (x))^{2} .

Join BYJU'S Learning Program

If y = sin (sin x) and d2ydx2+dydx tan x+f(x) = 0, then f(x) equals.

The correct option is D dydx=cos(sin x)cos x d2ydx2= I I=-cos(sin x)sin x−cos x[sin(sin x)]cos x I +dydxtan x=−cos2 x sin(sin x) d2ydx2+dydxtan x+cos2 x sin(sin x)=0 f(x)=cos2x sin(sin x)

If y = sin (sin x) and $\frac{d^{2} y}{d x^{2}} + \frac{d y}{d x}$ tan x+f(x) = 0, then f(x) equals.