If y=x4−10 and if x changes from 2 to 1.99, than the approximate change in y is

If y=x410 and if x changes from 2 to 1.99, than the approximate change in y is

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    Solution:

    Let x=2,x+Δx=1.99. Then, Δx=1.992=0.01.

    Let dx=Δx=0.01

    We have,

    y=x410dydx=4x3dydxx=2=4(2)3=32

    Now, dy=dydxdx

     dy=32(0.01)=0.32

     Δy=0.32 approximately [Δydy]

    So, approximate change in y=0.32.

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