Let y=2(x^2)+4x+3.

Find the differential dy when x=4 and dx=0.3

Find the differential dy when x=4 and dx=0.6

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The function y=2(x^2)+4x+3

dy = (4x + 4) dx

For x = 4 and dx = 0.3, dy = 20*0.3 = 6

For x = 4 and dx = 0.6, dy = 20*0.6 = 12

**The value of dy when x = 4 and dx = 0.3 is 6 and when x = 4 and dx = 0.6 dy = 12**.

You need to differentiate the function y with respect to x such that:

`(dy)/(dx) = (2x^2+4x+3)' =gt (dy)/(dx) = 4x + 4`

You need to multiply by dx both sides such that:

`(dy) = (4x + 4)(dx) `

You need to evaluate dy at x=4 and dx = 0.3, hence you should substitute 4 for x and 0.3 for dx such that:

`(dy) = (4*4 + 4)(0.3) `

`(dy) = (20)(0.3) =gt (dy) = 6`

You need to evaluate `dy` at x=4 and `dx = 0.6` , hence you should substitute 4 for x and 0.6 for dx such that:

`(dy) = (4*4 + 4)(0.6)`

`(dy) = (20)(0.6) =gt (dy) = 12`

**Hence, evaluating dy under given conditions yields `(dy) = 6 ` at `x=4` and `dx=0.3` ; `(dy) = 12` at `x=4` and `dx=0.6` .**

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