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Question

If xy+yx=ab the find dydx.  OR 
If ey(x+1)=1, then show that d2ydx2=(dydx)2.


Solution

We have xy+yx=abelogxy+elogyx=abeylogx+exlogy=ab

eylogx(y×1x+logx×dydx)+exlogy(x×1y×dydx+logy.1)=0

xy(yx+logx×dydx)+yx(xy×dydx+logy)=0

xy1y+xylogx×dydx+yx1x×dydx+yxlogy=0

(xylogx+yx1x)dydx=(yxlogy+xy1y)        dydx=(yxlogy+xy1yxylogx+yx1x)

OR We have ey(x+1)=1ey=1x+1logeey=loge(1x+1)or,y=log(x+1)

Now dydx=1x+1...(i) and, ddx(dydx)=ddx(1x+1)

d2ydx2=1(x+1)2=(1x+1)2=(dydx)2[By(i)

Therefore,   d2ydx2=(dydx)2.

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