# Derivatives of Elementary Functions

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Let's start with the simplest function, namely, the constant polynomial ${f{{\left({x}\right)}}}={c}$.

Derivative of a Constant Function. $\frac{{d}}{{{d}{x}}}{\left({c}\right)}={0}$.

Indeed, ${f{'}}{\left({x}\right)}=\lim_{{{h}\to{0}}}\frac{{{f{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}}}{{h}}=\lim_{{{h}\to{0}}}\frac{{{c}-{c}}}{{h}}=\lim_{{{h}\to{0}}}{0}={0}$.

Now, let's look at the power function ${f{{\left({x}\right)}}}={{x}}^{{n}}$.

Let's find a couple of derivatives for different values of ${n}$.

${n}={1}$: ${f{{\left({x}\right)}}}={x}$ and ${f{'}}{\left({x}\right)}=\lim_{{{h}\to{0}}}\frac{{{f{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}}}{{h}}=\lim_{{{h}\to{0}}}\frac{{{x}+{h}-{x}}}{{h}}=\lim_{{{h}\to{0}}}{1}={1}$.

${n}={2}$: ${f{{\left({x}\right)}}}={{x}}^{{2}}$ and ${f{'}}{\left({x}\right)}=\lim_{{{h}\to{0}}}\frac{{{f{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}}}{{h}}=\lim_{{{h}\to{0}}}\frac{{{{\left({x}+{h}\right)}}^{{2}}-{{x}}^{{2}}}}{{h}}=$

$=\lim_{{{h}\to{0}}}\frac{{{{x}}^{{2}}+{2}{x}{h}+{{h}}^{{2}}-{{x}}^{{2}}}}{{h}}=\lim_{{{h}\to{0}}}{\left({2}{x}+{h}\right)}={2}{x}+{0}={2}{x}$.

${n}={3}$: ${f{{\left({x}\right)}}}={{x}}^{{3}}$ and ${f{'}}{\left({x}\right)}=\lim_{{{h}\to{0}}}\frac{{{f{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}}}{{h}}=\lim_{{{h}\to{0}}}\frac{{{{\left({x}+{h}\right)}}^{{3}}-{{x}}^{{3}}}}{{h}}=\lim_{{{h}\to{0}}}{1}=$

$=\lim_{{{h}\to{0}}}\frac{{{{x}}^{{3}}+{3}{{x}}^{{2}}{h}+{3}{x}{{h}}^{{2}}+{{h}}^{{3}}-{{x}}^{{3}}}}{{h}}=\lim_{{{h}\to{0}}}{\left({3}{{x}}^{{2}}+{3}{x}{h}+{{h}}^{{2}}\right)}={3}{{x}}^{{2}}+{3}{x}\cdot{0}+{{0}}^{{2}}=$

$={3}{{x}}^{{2}}$.

There is a pattern. It seems that ${\left({{x}}^{{n}}\right)}'={n}{{x}}^{{{n}-{1}}}$ for positive ${n}$. It is true. And in general it can be proved using binomial theorem. But that's not all. We know that ${\left(\sqrt{{{x}}}\right)}'=\frac{{1}}{{{2}\sqrt{{{x}}}}}$ and it can be rewritten as ${\left({{x}}^{{\frac{{1}}{{2}}}}\right)}'=\frac{{1}}{{2}}{{x}}^{{\frac{{1}}{{2}}-{1}}}$. So, above rule holds even when ${n}$ is fraction.

It can also be proved when ${n}$ is negative or irrational.

Derivative of Power Function. If ${n}$ is any real number then ${\left({{x}}^{{n}}\right)}'={n}{{x}}^{{{n}-{1}}}$.

Example 1. Differentiate ${f{{\left({x}\right)}}}=\frac{{1}}{{{x}}^{{3}}}$.

Since $\frac{{1}}{{{x}}^{{3}}}={{x}}^{{-{3}}}$ then ${f{'}}{\left({x}\right)}=-{3}{{x}}^{{-{3}-{1}}}=-{3}{{x}}^{{-{4}}}=\frac{{3}}{{{x}}^{{4}}}$.

Example 2. Find equation of a tangent line to ${f{{\left({x}\right)}}}={\sqrt[{{3}}]{{{{x}}^{{5}}}}}$ at (1,1).

Since ${\sqrt[{{3}}]{{{{x}}^{{5}}}}}={{x}}^{{\frac{{5}}{{3}}}}$ then ${f{'}}{\left({x}\right)}=\frac{{5}}{{3}}{{x}}^{{\frac{{5}}{{3}}-{1}}}=\frac{{5}}{{3}}{{x}}^{{\frac{{2}}{{3}}}}=\frac{{5}}{{3}}{\sqrt[{{3}}]{{{{x}}^{{2}}}}}$.

Thus, ${f{'}}{\left({1}\right)}=\frac{{5}}{{3}}{\sqrt[{{3}}]{{{{1}}^{{2}}}}}=\frac{{5}}{{3}}$.

Therefore, equation of tangent line is ${y}-{1}=\frac{{5}}{{3}}{\left({x}-{1}\right)}$ or ${y}=\frac{{5}}{{3}}{x}-\frac{{2}}{{3}}$.

Now, let's try to find derivative of exponential function ${f{{\left({x}\right)}}}={{a}}^{{x}}$.

By definition, ${f{'}}{\left({x}\right)}=\lim_{{{h}\to{0}}}\frac{{{f{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}}}{{h}}=\lim_{{{h}\to{0}}}\frac{{{{a}}^{{{x}+{h}}}-{{a}}^{{x}}}}{{h}}=\lim_{{{h}\to{0}}}{{a}}^{{x}}\frac{{{{a}}^{{h}}-{1}}}{{h}}=$

$={{a}}^{{x}}\lim_{{{h}\to{0}}}\frac{{{{a}}^{{h}}-{1}}}{{h}}={{a}}^{{x}}{\ln{{\left({a}\right)}}}$. (See example 11 in Indeterminate Forms note for proof that $\lim_{{{h}\to{0}}}\frac{{{{a}}^{{h}}-{1}}}{{h}}={\ln{{\left({a}\right)}}}$).

Derivative of an Exponential Function. ${\left({{a}}^{{x}}\right)}'={{a}}^{{x}}{\ln{{\left({a}\right)}}}$.

In particular if ${a}={e}$, then ${\color{blue}{{{\left({{e}}^{{x}}\right)}'={{e}}^{{x}}}}}$.

Also note that for ${f{{\left({x}\right)}}}={{a}}^{{x}}$ we have that $\lim_{{{h}\to{0}}}\frac{{{{a}}^{{h}}-{1}}}{{h}}=\lim_{{{h}\to{0}}}\frac{{{{a}}^{{{0}+{h}}}-{{a}}^{{0}}}}{{h}}={f{'}}{\left({0}\right)}$.

So, ${f{'}}{\left({x}\right)}={f{'}}{\left({0}\right)}{{a}}^{{x}}$. This means that the rate of change of any exponential function is proportional to the function itself (the slope is proportional to the height).

Example 3. Find first derivative of ${f{{\left({x}\right)}}}={5}{{e}}^{{x}}-{2}\cdot{{3}}^{{x}}$.

Using the constant multiple and difference rules we obtain that

${f{'}}{\left({x}\right)}={\left({5}{{e}}^{{x}}-{2}\cdot{{3}}^{{x}}\right)}'={\left({5}{{e}}^{{x}}\right)}'-{\left({2}\cdot{{3}}^{{x}}\right)}'={5}{\left({{e}}^{{x}}\right)}'-{2}{\left({{3}}^{{x}}\right)}'={5}{{e}}^{{x}}-{2}{\ln{{\left({3}\right)}}}{{3}}^{{x}}$.

Now, it is time to find derivative of logarithmic function ${f{{\left({x}\right)}}}={\log}_{{a}}{\left({x}\right)}$.

By definition, $\lim_{{{h}\to{0}}}\frac{{{f{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}}}{{h}}=\lim_{{{h}\to{0}}}\frac{{{\log}_{{a}}{\left({x}+{h}\right)}-{\log}_{{a}}{\left({x}\right)}}}{{h}}=\lim_{{{h}\to{0}}}\frac{{{\log}_{{a}}{\left(\frac{{{x}+{h}}}{{x}}\right)}}}{{h}}=\lim_{{{h}\to{0}}}\frac{{{\log}_{{a}}{\left({1}+\frac{{h}}{{x}}\right)}}}{{h}}=$

$=\lim_{{{h}\to{0}}}\frac{{1}}{{x}}\frac{{x}}{{h}}{\log}_{{a}}{\left({1}+\frac{{h}}{{x}}\right)}=\lim_{{{h}\to{0}}}\frac{{1}}{{x}}{\log}_{{a}}{\left({{\left({1}+\frac{{h}}{{x}}\right)}}^{{\frac{{x}}{{h}}}}\right)}$.

Now, if we denote $\frac{{x}}{{h}}$ as ${t}$ then for fixed ${x}$ we have that ${t}\to\infty$ as ${h}\to{0}$.

So, $\lim_{{{h}\to{0}}}\frac{{1}}{{x}}{\log}_{{a}}{\left({{\left({1}+\frac{{h}}{{x}}\right)}}^{{\frac{{x}}{{h}}}}\right)}=\lim_{{{t}\to\infty}}\frac{{1}}{{x}}{\log}_{{a}}{\left({{\left({1}+\frac{{1}}{{t}}\right)}}^{{t}}\right)}=\frac{{1}}{{x}}{\log}_{{a}}{\left(\lim_{{{t}\to\infty}}{{\left({1}+\frac{{1}}{{t}}\right)}}^{{t}}\right)}=\frac{{1}}{{x}}{\log}_{{a}}{\left({e}\right)}=$

$=\frac{{1}}{{x}}\frac{{{\ln{{\left({e}\right)}}}}}{{\ln{{\left({a}\right)}}}}=\frac{{1}}{{{x}{\ln{{\left({a}\right)}}}}}$.

Derivative of Logarithmic function. ${\log}_{{a}}{\left({x}\right)}=\frac{{1}}{{{x}{\ln{{\left({a}\right)}}}}}$.

In particular if ${a}={e}$ then ${\left({\ln{{\left({x}\right)}}}\right)}'=\frac{{1}}{{x}}$.

Now let's find derivative of trigonometric functions.

Let ${y}={\sin{{\left({x}\right)}}}$.

By definition ${f{'}}{\left({x}\right)}=\lim_{{{h}\to{0}}}\frac{{{\sin{{\left({x}+{h}\right)}}}-{\sin{{\left({x}\right)}}}}}{{h}}$.

Here we need to convert a difference of sines into product: ${\sin{{\left({x}+{h}\right)}}}-{\sin{{\left({x}\right)}}}={2}{\sin{{\left(\frac{{{x}+{h}-{x}}}{{2}}\right)}}}{\cos{{\left(\frac{{{x}+{h}+{x}}}{{2}}\right)}}}={2}{\sin{{\left(\frac{{h}}{{2}}\right)}}}{\cos{{\left({x}+\frac{{h}}{{2}}\right)}}}$.

Now, the limit can be rewritten as $\lim_{{{h}\to{0}}}\frac{{{2}{\sin{{\left(\frac{{h}}{{2}}\right)}}}{\cos{{\left({x}+\frac{{h}}{{2}}\right)}}}}}{{h}}=\lim_{{{h}\to{0}}}\frac{{{\sin{{\left(\frac{{h}}{{2}}\right)}}}}}{{\frac{{h}}{{2}}}}{\cos{{\left({x}+\frac{{h}}{{2}}\right)}}}=$

$=\lim_{{{h}\to{0}}}\frac{{{\sin{{\left(\frac{{h}}{{2}}\right)}}}}}{{\frac{{h}}{{2}}}}\lim_{{{h}\to{0}}}{\cos{{\left({x}+\frac{{h}}{{2}}\right)}}}={1}\cdot{\cos{{\left({x}+\frac{{0}}{{2}}\right)}}}={\cos{{\left({x}\right)}}}.$

Derivative of Sine. ${\left({\sin{{\left({x}\right)}}}\right)}'={\cos{{\left({x}\right)}}}$.

Similarly can be found that ${\left({\cos{{\left({x}\right)}}}\right)}'=-{\sin{{\left({x}\right)}}}$.

Derivative of Cosine. ${\left({\cos{{\left({x}\right)}}}\right)}'=-{\sin{{\left({x}\right)}}}$.

We can find derivative of tangent using definition, but it is simpler to use the Quotient Rule:

${\left({\tan{{\left({x}\right)}}}\right)}'={\left(\frac{{{\sin{{\left({x}\right)}}}}}{{{\cos{{\left({x}\right)}}}}}\right)}'=\frac{{{\left({\sin{{\left({x}\right)}}}\right)}'{\cos{{\left({x}\right)}}}-{\sin{{\left({x}\right)}}}{\left({\cos{{\left({x}\right)}}}\right)}'}}{{{{\cos}}^{{2}}{\left({x}\right)}}}=\frac{{{\cos{{\left({x}\right)}}}{\cos{{\left({x}\right)}}}-{\sin{{\left({x}\right)}}}{\left(-{\sin{{\left({x}\right)}}}\right)}}}{{{{\cos}}^{{2}}{\left({x}\right)}}}=$

$=\frac{{{{\cos}}^{{2}}{\left({x}\right)}+{{\sin}}^{{2}}{\left({x}\right)}}}{{{{\cos}}^{{2}}{\left({x}\right)}}}=\frac{{1}}{{{{\cos}}^{{2}}{\left({x}\right)}}}={{\sec}}^{{2}}{\left({x}\right)}$.

Derivative of Tangent. ${\left({\tan{{\left({x}\right)}}}\right)}'=\frac{{1}}{{{{\cos}}^{{2}}{\left({x}\right)}}}={{\sec}}^{{2}}{\left({x}\right)}$.

Similarly, we can find that derivative of cotangent is ${\left({\cot{{\left({x}\right)}}}\right)}'=-{{\csc}}^{{2}}{\left({x}\right)}$.

Using quotient rule we can find derivatives of secant and cosecant.

For example. ${\left({\sec{{\left({x}\right)}}}\right)}'={\left(\frac{{1}}{{\cos{{\left({x}\right)}}}}\right)}'=\frac{{{\left({1}\right)}'{\cos{{\left({x}\right)}}}-{1}\cdot{\left({\cos{{\left({x}\right)}}}\right)}'}}{{{{\cos}}^{{2}}{\left({x}\right)}}}=\frac{{{0}\cdot{\cos{{\left({x}\right)}}}+{\sin{{\left({x}\right)}}}}}{{{{\cos}}^{{2}}{\left({x}\right)}}}=\frac{{{\sin{{\left({x}\right)}}}}}{{{{\cos}}^{{2}}{\left({x}\right)}}}=$

$=\frac{{{\sin{{\left({x}\right)}}}}}{{{\cos{{\left({x}\right)}}}}}\cdot\frac{{1}}{{\cos{{\left({x}\right)}}}}={\tan{{\left({x}\right)}}}{\sec{{\left({x}\right)}}}$.

Derivative of Secant. ${\left({\sec{{\left({x}\right)}}}\right)}'={\tan{{\left({x}\right)}}}{\sec{{\left({x}\right)}}}$.

Derivative of Cosecant. ${\left({\csc{{\left({x}\right)}}}\right)}'=-{\cot{{\left({x}\right)}}}{\csc{{\left({x}\right)}}}$.

To find derivative of inverse trigonometric functions it is simpler to use implicit differentiation.

By definition if ${y}={\operatorname{arcsin}{{\left({x}\right)}}}$ then ${\sin{{\left({y}\right)}}}={x}$ for $-\frac{\pi}{{2}}\le{y}\le\frac{\pi}{{2}}$.

Now differentiate ${\sin{{\left({y}\right)}}}={x}$ with respect to ${x}$: ${\cos{{\left({y}\right)}}}\cdot{y}'={x}'$ or ${\cos{{\left({y}\right)}}}{y}'={1}$.

From this we have that ${y}'=\frac{{1}}{{{\cos{{\left({y}\right)}}}}}$. Now we need to express this in terms of ${x}$.

From identity ${{\cos}}^{{2}}{\left({y}\right)}+{{\sin}}^{{2}}{\left({y}\right)}={1}$ we have that ${{\cos}}^{{2}}{\left({y}\right)}={1}-{{\sin}}^{{2}}{\left({y}\right)}={1}-{{x}}^{{2}}$. Since ${y}$ is changing in interval ${\left[-\frac{\pi}{{2}},\frac{\pi}{{2}}\right]}$ then cosine should pe positive. This means that ${\cos{{\left({y}\right)}}}=\sqrt{{{1}-{{x}}^{{2}}}}$.

Derivative of Inverse Sine. ${\left({\operatorname{arcsin}{{\left({x}\right)}}}\right)}'=\frac{{1}}{{\sqrt{{{1}-{{x}}^{{2}}}}}}$.

Similarly can be found other derivatives of inverse trigonometric functions.

Derivatives of Hyperbolic Functions can be easily found using constant multiply, sum, quotient and chain rules, because they are rationally expresses through exponent.

For example, ${\left({\sinh{{\left({x}\right)}}}\right)}'={\left(\frac{{{{e}}^{{x}}-{{e}}^{{-{x}}}}}{{2}}\right)}'=\frac{{1}}{{2}}{\left({{e}}^{{x}}-{{e}}^{{-{x}}}\right)}'=\frac{{1}}{{2}}{\left({\left({{e}}^{{x}}\right)}'-{\left({{e}}^{{-{x}}}\right)}'\right)}=\frac{{1}}{{2}}{\left({{e}}^{{x}}-{{e}}^{{-{x}}}\cdot{\left(-{x}\right)}'\right)}=$

$=\frac{{1}}{{2}}{\left({{e}}^{{x}}+{{e}}^{{-{x}}}\right)}={\cosh{{\left({x}\right)}}}$.

Derivative of Hyperbolic Sine. ${\left({\sinh{{\left({x}\right)}}}\right)}'={\cosh{{\left({x}\right)}}}$.

Inverse Hyperbolic Functions can be found using standard differentiation rules.

For example, inverse hyperbolic cosine can be found using the chain and sum rules.

${\left(\text{arccosh}{\left({x}\right)}\right)}'={\left({\ln{{\left({x}+\sqrt{{{{x}}^{{2}}-{1}}}\right)}}}\right)}'=\frac{{1}}{{{x}+\sqrt{{{{x}}^{{2}}-{1}}}}}\cdot{\left({x}+\sqrt{{{{x}}^{{2}}+{1}}}\right)}'=\frac{{{1}+\frac{{1}}{{{2}\sqrt{{{{x}}^{{2}}+{1}}}}}\cdot{\left({{x}}^{{2}}+{1}\right)}'}}{{{x}+\sqrt{{{{x}}^{{2}}+{1}}}}}=$

$=\frac{{{1}+\frac{{x}}{{\sqrt{{{{x}}^{{2}}+{1}}}}}}}{{{x}+\sqrt{{{{x}}^{{2}}+{1}}}}}=\frac{{\frac{{{x}+\sqrt{{{{x}}^{{2}}+{1}}}}}{{\sqrt{{{{x}}^{{2}}+{1}}}}}}}{{{x}+\sqrt{{{{x}}^{{2}}+{1}}}}}=\frac{{1}}{{\sqrt{{{{x}}^{{2}}+{1}}}}}$.

Derivative of Inverse Hyperbolic Cosine. ${\left(\text{arccosh}{\left({x}\right)}\right)}'=\frac{{1}}{{\sqrt{{{{x}}^{{2}}+{1}}}}}$.

Derivatives of elementary functions can be found in the table of derivatives.