Using differential, find the approximate value of `f(2.01)` where `f(x)=4x^3+5x^2+2`

Explanation

`f(x)=f(a)+(x-a)f'(a)`

`f(x)=4x^2+5x^2+2`

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

`f(2.01)=f(2)+(2.01-2)f'(2)`

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

`f'(x)=68`

`f(2.01)=54+0.01*68=54.68`.