First slide
Derivatives
Question

if,y=xlogxa+bx then x3d2ydx2=

Moderate
Solution

 yx=logxlog(a+bx)

xdydxyx2=1xba+bx=ax(a+bx)xdydxy=axa+bx...(1)

Diff again w..r.t ‘x’ we get

xd2ydx2+dydxdydx=(a+bx)aaxb(a+bx)2xd2ydx2=a2(a+bx)2x3d2ydx2=a2x2(a+bx)2=xdydxy2by eq(1)

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