If y=5x^2+6x+6,x=2,and Delta x=0.001,then the | vedclass

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(B) Given $y = f(x) = 5x^2 + 6x + 6$,$x = 2$,and $\Delta x = 0.001$.First,calculate the differential $dy$:$\frac{dy}{dx} = 10x + 6$$dy = \left(\frac{d

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$0.026005$ and $0.026$

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(B) Given $y = f(x) = 5x^2 + 6x + 6$,$x = 2$,and $\Delta x = 0.001$.First,calculate the differential $dy$:$\frac{dy}{dx} = 10x + 6$$dy = \left(\frac{dy}{dx}\right) \Delta x = (10x + 6) \Delta x$Substituting $x = 2$ and $\Delta x = 0.001$:$dy = (10(2) + 6)(0.001) = (26)(0.001) = 0.026$.Next,calculate the increment $\Delta y$:$\Delta y = f(x + \Delta x) - f(x)$$\Delta y = [5(x + \Delta x)^2 + 6(x + \Delta x) + 6] - [5x^2 + 6x + 6]$$\Delta y = 5(x^2 + 2x\Delta x + (\Delta x)^2) + 6x + 6\Delta x + 6 - 5x^2 - 6x - 6$$\Delta y = 10x\Delta x + 5(\Delta x)^2 + 6\Delta x$Substituting $x = 2$ and $\Delta x = 0.001$:$\Delta y = 10(2)(0.001) + 5(0.001)^2 + 6(0.001)$$\Delta y = 0.020 + 0.000005 + 0.006 = 0.026005$.Thus,$\Delta y = 0.026005$ and $dy = 0.026$.

If $y=5x^2+6x+6$,$x=2$,and $\Delta x=0.001$,then the values of $\Delta y$ and $dy$ respectively are:

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