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3.33 Acceleration as the second derivative

So, taking the limit as h→0h\rightarrow 0,

f′⁢(x)=limh→0⁡f⁢(x+h)-f⁢(x)h=(velocity at time x);f^{\prime}(x)=\lim_{h\rightarrow 0}{{f(x+h)-f(x)}\over{h}}=({\hbox{velocity at time}}\quad x);
f′′⁢(x)=(acceleration at time x).f^{\prime\prime}(x)=({\hbox{acceleration at time}}\quad x).

Example

To find the higher order derivatives of y=xny=x^{n}.

Solution. We compute the derivatives in a column to help us recognise patterns: