# Can two functions have identical first and second derivatives, but different higher order derivatives?

Consider two functions $f(x)$ and $g(x)$ defined for all $x\in\mathbb{R}$. Assume that $\forall x\in\mathbb{R}$, $f(x)=g(x)$, $f'(x)=g'(x)$, and $f''(x)=g''(x)$. Is it possible that higher order derivatives of $f(x)$ and $g(x)$ are not equal?

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why did you include the assumption $f(x) = g(x)$? It sounds like you are interested in the case when $f(x) \neq g(x)$... – algebra_fan Jul 22 '11 at 23:27

Since the higher order derivatives are given simply by iteration of the "simple" derivative the answer is no. That is to say, if $f^{(n)}(x) = g^{(n)}(x)$ then $f^{(n+1)}(x) = \left( \, f^{(n)}(x)\right)' = \left(g^{(n)}(x)\right)' = g^{(n+1)}(x)$. Excuse the abuse of notation.