wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

x+y+yx=c, then d2ydx2

Open in App
Solution

We have,

x+y+yx=c ……… (1)

On differentiating above equation w.r.t x, we get

12x+y×(1+dydx)+12yx×(dydx1)=0

12x+y+dydx12x+y+12yxdydx12yx=0

12(1x+y+1yx)dydx=12(1yx1x+y)

(yx+x+yx+yyx)dydx=(x+yyxx+yyx)

dydx=(x+yyxx+y+yx)

dydx=(x+yyxx+y+yx)×(x+yyxx+yyx)

dydx=(x+yyx)2(x+y)2(yx)2

dydx=(x+y+yx2x+y×yx)x+yy+x

dydx=(2y2y+x×yx)2x

dydx=(yy2x2)x ……… (2)

On differentiating above equation w.r.t x, we get

d2ydx2=x(dydx(12y2x2×(2ydydx2x)))(yy2x2)×1x2

d2ydx2=x(dydx(1y2x2×(ydydxx)))(yy2x2)x2

From equation (2), we get

d2ydx2=x⎜ ⎜⎜ ⎜(yy2x2)x⎟ ⎟⎜ ⎜1y2x2×⎜ ⎜y⎜ ⎜(yy2x2)x⎟ ⎟x⎟ ⎟⎟ ⎟⎟ ⎟(yy2x2)x2

d2ydx2=⎜ ⎜(yy2x2)⎜ ⎜xy2x2×⎜ ⎜y(yy2x2)x2x⎟ ⎟⎟ ⎟⎟ ⎟(yy2x2)x2

d2ydx2=(yy2x2)⎜ ⎜y(yy2x2)x2y2x2⎟ ⎟(yy2x2)x2

d2ydx2=⎜ ⎜(y2yy2x2)x2y2x2⎟ ⎟x2

d2ydx2=x2y2+yy2x2x2y2x2

Hence, this is the answer.


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Basic Theorems in Differentiation
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon