Derivatives of Elementary Functions

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 a positive ${n}$. It is true. And, in general, it can be proved using the 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, the above rule holds even when ${n}$ is a fraction.

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

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

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

Since $\frac{{1}}{{{x}}^{{3}}}={{x}}^{{-{3}}}$, it can be stated that ${f{'}}{\left({x}\right)}=-{3}{{x}}^{{-{3}-{1}}}=-{3}{{x}}^{{-{4}}}=\frac{{3}}{{{x}}^{{4}}}$.

Let's move on to another example.

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

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

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

Therefore, the equation of the 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 the derivative of the 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 the Indeterminate Forms note for the 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 ${a}={e}$, it can be stated that ${\color{blue}{{{\left({{e}}^{{x}}\right)}'={{e}}^{{x}}}}}$.

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 the 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 the derivative of the 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 ${t}$, for the 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 a Logarithmic Function. ${\log}_{{a}}{\left({x}\right)}=\frac{{1}}{{{x}{\ln{{\left({a}\right)}}}}}$.

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

Now, let's find the derivatives 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 the difference of sines into a 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, it 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 the derivative of tangent using the 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 the derivative of cotangent is ${\left({\cot{{\left({x}\right)}}}\right)}'=-{{\csc}}^{{2}}{\left({x}\right)}$.

Using the quotient rule, we can find the 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 the derivatives of inverse trigonometric functions, it is simpler to use implicit differentiation.

By definition, if ${y}={\operatorname{asin}{{\left({x}\right)}}}$, it can be stated that ${\sin{{\left({y}\right)}}}={x}$ 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 the 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 the interval ${\left[-\frac{\pi}{{2}},\frac{\pi}{{2}}\right]}$, the cosine should be positive. This means that ${\cos{{\left({y}\right)}}}=\sqrt{{{1}-{{x}}^{{2}}}}$.

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

The other derivatives of inverse trigonometric functions can be found similarly.

The derivatives of hyperbolic functions can be found easily using the constant multiply, sum, quotient, and chain rules, because they are rationally expressed through the 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, the inverse hyperbolic cosine can be found using the chain and sum rules:

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

$=\frac{{{1}+\frac{{1}}{{{2}\sqrt{{{{x}}^{{2}}+{1}}}}}\cdot{\left({{x}}^{{2}}+{1}\right)}'}}{{{x}+\sqrt{{{{x}}^{{2}}+{1}}}}}=\frac{{{1}+\frac{{x}}{{\sqrt{{{{x}}^{{2}}+{1}}}}}}}{{{x}+\sqrt{{{{x}}^{{2}}+{1}}}}}=\frac{{\frac{{{x}+\sqrt{{{{x}}^{{2}}+{1}}}}}{{\sqrt{{{{x}}^{{2}}+{1}}}}}}}{{{x}+\sqrt{{{{x}}^{{2}}+{1}}}}}=\frac{{1}}{{\sqrt{{{{x}}^{{2}}+{1}}}}}$.

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

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