Using differentials, find the approximate value of `f(2. 01),` where `f(x)=4x^3+5x^2+2.`

Explanation

Here, we will use the definition of derivatives:

`f'(x) = (f(x+Deltax)-f(x))/((x+Deltax)-x)`

`f'(x) = (f(x+Deltax)-f(x))/(Deltax)`

Here, `x = 2 and Deltax = 0.01`

`:. f'(2) = (f(2+0.01) - f(2))/0.01->(1)`

Here, `f'(x) = 12x^2+10x`

`:. f'(2) = 12(4)+10(2) = 68`

`f(2) = 4(8)+5(4)+2 = 54 `

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