# Formulas for Higher-Order Derivatives

In general, to find n-th derivative of function $y={f{{\left({x}\right)}}}$ we need to find all derivatives of previous orders. But sometimes it is possible to obtain expression for n-th derivative that depends on $n$ and doesn't contain previous derivatives.

First of all it is easily to extend constant multiple, sum and difference rules.

Fact. For functions ${f{{\left({x}\right)}}}$ and ${g{{\left({x}\right)}}}$, and constant ${c}$ we have that

• ${{\left({c}{f}\right)}}^{{{\left({n}\right)}}}={c}\cdot{{f}}^{{{\left({n}\right)}}}$;
• ${{\left({f{\pm}}{g}\right)}}^{{{\left({n}\right)}}}={{f}}^{{{\left({n}\right)}}}\pm{{g}}^{{{\left({n}\right)}}}$.

Now consider power function ${y}={{x}}^{{a}}$.

We have that ${y}'={a}{{x}}^{{{a}-{1}}}$, ${y}''={a}{\left({a}-{1}\right)}{{x}}^{{{a}-{2}}}$.

In general, ${\color{red}{{{{\left({{x}}^{{a}}\right)}}^{{{\left({n}\right)}}}={a}{\left({a}-{1}\right)}\ldots{\left({a}-{n}+{1}\right)}{{x}}^{{{a}-{n}}}}}}$.

Example 1. Find ${{y}}^{{{\left({7}\right)}}}$ if ${y}={2}{{x}}^{{5}}-{5}{{x}}^{{10}}$.

We first use constant multiple and sum rule and then formula for power function.

${{y}}^{{{\left({7}\right)}}}={{\left({2}{{x}}^{{5}}-{5}{{x}}^{{10}}\right)}}^{{{\left({7}\right)}}}={2}{{\left({{x}}^{{5}}\right)}}^{{{\left({7}\right)}}}-{5}{{\left({{x}}^{{10}}\right)}}^{{{\left({7}\right)}}}=$

$={2}{\left({5}\cdot{4}\cdot{3}\cdot{2}\cdot{1}\cdot{0}\cdot{\left(-{1}\right)}\cdot{{x}}^{{{5}-{7}}}\right)}-{5}{\left({10}\cdot{9}\cdot{8}\cdot{7}\cdot{6}\cdot{5}\cdot{4}\cdot{{x}}^{{{10}-{7}}}\right)}=-{3024000}{{x}}^{{3}}$.

For function ${y}={{a}}^{{x}}$ we have that ${y}'={{a}}^{{x}}{\ln{{\left({a}\right)}}}$, ${y}''={{a}}^{{x}}{{\left({\ln{{\left({a}\right)}}}\right)}}^{{2}}$ and in general ${\color{green}{{{{\left({{a}}^{{x}}\right)}}^{{{\left({n}\right)}}}={{a}}^{{x}}{{\left({\ln{{\left({a}\right)}}}\right)}}^{{n}}}}}$. In particular, ${\color{blue}{{{{\left({{e}}^{{x}}\right)}}^{{{\left({n}\right)}}}={{e}}^{{x}}}}}$.

For function ${y}={\sin{{\left({x}\right)}}}$ we have that ${y}'={\cos{{\left({x}\right)}}}$, ${y}''=-{\sin{{\left({x}\right)}}}$, ${y}'''=-{\cos{{\left({x}\right)}}}$, ${{y}}^{{{\left({4}\right)}}}={\sin{{\left({x}\right)}}}$.

Since ${\cos{{\left({x}\right)}}}={\sin{{\left({x}+\frac{\pi}{{2}}\right)}}}$ then $\color{green} {{{{\left({\sin{{\left({x}\right)}}}\right)}}^{{{\left({n}\right)}}}={\sin{{\left({x}+{n}\frac{\pi}{{2}}\right)}}}}}$.

Similarly, $\color{purple} {{{{\left({\cos{{\left({x}\right)}}}\right)}}^{{{\left({n}\right)}}}={\cos{{\left({x}+{n}\frac{\pi}{{2}}\right)}}}}}$.

We saw that constant multiple and sum rules are easily extended for n-th derivative.

It is not a case with product rule.

So, let's see how product rule looks like in the case of higher-order derivatives.

Suppose that function ${f{{\left({x}\right)}}}$ and ${g{{\left({x}\right)}}}$ has finite derivatives up to n-th order.

We have that ${\left({f{{g}}}\right)}'={f{'}}{g{+}}{f{{g{'}}}}$, ${\left({f{{g}}}\right)}''={\left({f{'}}{g{+}}{f{{g{'}}}}\right)}'={\left({f{'}}{g}\right)}'+{\left({f{{g{'}}}}\right)}'={\left({f{''}}{g{+}}{f{'}}{g{'}}\right)}+{\left({f{'}}{g{'}}+{f{{g{''}}}}\right)}={f{''}}{g{+}}{2}{f{'}}{g{'}}+{f{{g{''}}}}$.

For third derivative we have ${\left({f{{g}}}\right)}'''={\left({f{''}}{g{+}}{2}{f{'}}{g{'}}+{f{{g{''}}}}\right)}'={\left({f{''}}{g}\right)}'+{2}{\left({f{'}}{g{'}}\right)}'+{\left({f{{g{''}}}}\right)}'=$

$={\left({f{'''}}{g{+}}{f{''}}{g{'}}\right)}+{2}{\left({f{''}}{g{'}}+{f{'}}{g{''}}\right)}+{\left({f{'}}{g{''}}+{f{{g{'''}}}}\right)}={f{'''}}{g{+}}{3}{f{''}}{g{+}}{3}{f{'}}{g{''}}+{f{{g{'''}}}}$.

We see that these formulas are very similar to the expansion of binom of newton. Just instead of powers here are derivatives. If we denote ${f{=}}{{f}}^{{{\left({0}\right)}}}$ and ${g{=}}{{g}}^{{{\left({0}\right)}}}$ then we will obtain Leibniz Formula.

Leibniz Formula. ${{\left({f{{g}}}\right)}}^{{{\left({n}\right)}}}={\sum_{{{i}={0}}}^{{n}}}{{C}_{{n}}^{{i}}}{{f}}^{{{\left({n}-{i}\right)}}}{{g}}^{{{\left({i}\right)}}}$.

We can extend this case when in product there are more than two functions, but this involves many formulas and calculations fairly complex.

Example 2. Find ${{\left({{x}}^{{2}}{\cos{{\left({x}\right)}}}\right)}}^{{{\left({50}\right)}}}$.

Here ${g{{\left({x}\right)}}}={{x}}^{{2}}$ and ${f{{\left({x}\right)}}}={\cos{{\left({x}\right)}}}$.

We have that, ${{\left({f{{\left({x}\right)}}}\right)}}^{{{\left({k}\right)}}}={\cos{{\left({x}+{k}\frac{\pi}{{2}}\right)}}}$. Also ${g{'}}{\left({x}\right)}={2}{x}$, ${g{''}}{\left({x}\right)}={2}$, ${g{'''}}{\left({x}\right)}={{g}}^{{{\left({4}\right)}}}{\left({x}\right)}=\ldots={0}$.

This means that in Leibniz formula all summands except first three will be equal 0. Thus, ${{\left({f{{g}}}\right)}}^{{{\left({50}\right)}}}={{C}_{{50}}^{{0}}}{{f}}^{{{\left({50}\right)}}}{g{+}}{{C}_{{50}}^{{1}}}{{f}}^{{{\left({49}\right)}}}{g{'}}+{{C}_{{50}}^{{2}}}{{f}}^{{{\left({48}\right)}}}{g{''}}=$

$={1}\cdot{\cos{{\left({x}+{50}\frac{\pi}{{2}}\right)}}}\cdot{{x}}^{{2}}+{50}\cdot{\cos{{\left({x}+{49}\frac{\pi}{{2}}\right)}}}\cdot{2}{x}+\frac{{{50}\cdot{49}}}{{2}}\cdot{\cos{{\left({x}+{48}\frac{\pi}{{2}}\right)}}}\cdot{2}=$

$=-{{x}}^{{2}}{\cos{{\left({x}\right)}}}-{100}{x}{\sin{{\left({x}\right)}}}+{2450}{\cos{{\left({x}\right)}}}={\left({2450}-{{x}}^{{2}}\right)}{\cos{{\left({x}\right)}}}-{100}{x}{\sin{{\left({x}\right)}}}$.

So, ${{\left({{x}}^{{2}}{\cos{{\left({x}\right)}}}\right)}}^{{{\left({50}\right)}}}={\left({2450}-{{x}}^{{2}}\right)}{\cos{{\left({x}\right)}}}-{100}{x}{\sin{{\left({x}\right)}}}$.