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Question

If loge(x+y)=4xy. Find d2ydx2 at x=0


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Solution

Step 1: Solve the value of y when x=0

loge(x+y)=4xy

x+y=e4xy ...(i) logea=beb=a

Substitute value of x=0

y=e0

y=1

Step 2: Solve for the first order derivative of given equation

Differentiating with respect to x we get

1+dydx=e4xy4xdydx+4y ddxefx,y=efx,y×ddxfx,y

dydx1-4xe4xy=4ye4xy-1

dydx=4ye4xy-11-4xe4xy

Substitute the values of x=0,y=1

dydx=41-11-0

dydx=3

Step 3 : Solve for the second order derivative

dydx=4ye4xy-11-4xe4xy

dydx=4yx+y-11-4xx+y [From (i)]

dydx=4y2+4xy-11-4x2-4xy ...(ii)

Differentiating ii with respect to x we get

d2ydx2=1-4x2-4xy8ydydx+4y+4xdydx-4y2+4xy-1-8x-4y-4xdydx1-4x2-4xy2 ddxuv=vdudx-udvdxv2

d2ydx2=1-4x2-4xy(4x+8y)dydx+4y+4y2+4xy-18x+4y+4xdydx1-4x2-4xy2

Substitute the value of x=0,y=1,dydx=3

d2ydx2=1-0-0(0+8)(3)+4+4+0-10+4+01-0-02

d2ydx2=28+3412

d2ydx2=40

Hence, the value of d2ydx2 at x=0 is 40.


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