The derivatives of standard functions

The constant function\(: (\lambda )' \)

A constant function is defined as followed:

With the definition of derivatives , we do have:

So,

In this case, there is no indeterminate form "\( \frac{0}{0} \)" because only the denominator goes to \(0 \), the numerator remains \(0 \) as a constant.

We definitely have the number \( 0 \) which is theorically divided by a number approaching \(0 \), which is still worth \(0 \).

The affine function\( : (ax + b )' \)

An affine function is defined as followed:

With the definition of derivatives , we do have:

So,

The absolute value function\(: |x|' \)

The absolute value function is defined as followed:

By using the derivatives of composite functions , we do have:

So,

The square function\(: (x^2 )' \)

The square function is defined as followed:

With the definition of derivatives , we do have:

So,

The powers of x functions\(: (x^n)' \)

In this part, there will be many cases where \(x\) will be found in the denominator, then for simplicity we will remove the case where \((x = 0)\).

Then, we specifically define the power of \(x\) function as follows:

We will make the demonstration of the general formula for all the successive sets.

1. With an exponent \( n \) in the natural numbers set \( (n \in \mathbb{N})\)

   With the definition of derivatives , we do have:

   $$ (x^n)' = \lim_{h \to 0 } \enspace \frac{(x + h)^n - x^n}{h} \qquad (1)$$

   But we know from the Newton's binomial formula that:

   $$\forall n \in \mathbb{N}, \enspace \forall (a, b) \in \hspace{0.04em} \mathbb{R}^2,$$

   $$ (a + b)^n = \sum_{p = 0}^n \binom{n}{p} a^{n-p}b^p $$

   Let us replace \( (a + b) \) by \( (x + h) \) to fit with the formula \( (1) \):

   $$ (x + h)^n = \sum_{p = 0}^n \binom{n}{p} x^{n-p}h^p$$

   $$ (x + h)^n = x^n + \binom{n}{1}x^{n - 1}h + \binom{n}{2}x^{n - 2}h^2 \enspace + ... + \enspace \binom{n}{n- 1}xh^{n - 1 } + h^{n} \qquad (2) $$

   Injecting \( (2) \) into \( (1) \), we do have now:

   $$ (x^n)' = \lim_{h \to 0 } \enspace \frac{x^n + \binom{n}{1}x^{n - 1}h + \binom{n}{2}x^{n - 2}h^2 \enspace + ... + \enspace \binom{n}{n- 1}xh^{n - 1 } + h^{n} - x^n}{h} $$

   Both terms \( x^n \) cancelled at the numerator:

   $$ (x^n)' = \lim_{h \to 0 } \enspace \frac{\binom{n}{1}x^{n - 1}h + \binom{n}{2}x^{n - 2}h^2 \enspace + ... + \enspace \binom{n}{n- 1}xh^{n - 1 } + h^{n}}{h} $$

   Now, we can simplify by \( h \) :

   $$ (x^n)' = \lim_{h \to 0 } \enspace \binom{n}{1}x^{n - 1} + \binom{n}{2}x^{n - 2}h \enspace + ... + \enspace \binom{n}{n- 1}xh^{n - 2 } + h^{n - 1} $$

   Taking the limit when \( h \to 0 \), all terms \( h \) disappear:

   $$ (x^n)' = \binom{n}{1}x^{n - 1} $$

   And finally,

   $$ \forall x \in \mathbb{R^*}, \enspace \forall n \in \mathbb{N}, $$
   $$ (x^n)' = nx^{n - 1} $$

2. With an exponent \( n \) in the integers set \( (n \in \mathbb{Z})\)

   We saw above the case of an integer positif exponent \( n \), now let's see the other part of : the \( \mathbb{Z}\) set, the negative integers.

   We then consider the set of negative integers only \( \mathbb{Z} \hspace{0.2em} \backslash \hspace{0.2em} \bigl \{ \mathbb{N} \bigr \} \), and a number \( n \) belonging to this set such that:

   $$ \forall m \in \mathbb{N}, \ n = -m $$

   Using the same method as above from the the definition of the derivative :

   $$ (x^n)' = \lim_{h \to 0 } \enspace \frac{(x + h)^n - x^n}{h} $$
   $$ (x^n)' = \lim_{h \to 0 } \enspace \frac{(x + h)^{-m} - x^{-m}}{h} $$
   $$ (x^n)' = \lim_{h \to 0 } \enspace \frac{\frac{1}{(x + h)^{m}} - \frac{1}{x^{m}} }{h} $$

   We put the big numerator under the same denominator.

   $$ (x^n)' = \lim_{h \to 0 } \enspace \frac{\frac{x^m}{(x + h)^{m}x^{m}} - \frac{(x + h)^{m}}{x^{m}(x + h)^{m}} }{h} $$
   $$ (x^n)' = \lim_{h \to 0 } \enspace \frac{\frac{x^m - (x + h)^{m}}{(x + h)^{m}x^{m}} }{h} $$
   $$ (x^n)' = \lim_{h \to 0 } \enspace \frac{1}{h} \Biggl[ \frac{x^m - (x + h)^{m}}{(x + h)^{m}x^{m}} \Biggr] $$

   But we already carried out this calculation above, and after simplification, we had:

   $$ \frac{ (x + h)^n - x^n }{h} = \binom{n}{1}x^{n - 1} + \binom{n}{2}x^{n - 2}h \enspace + ... + \enspace \binom{n}{n- 1}xh^{n - 2 } + h^{n - 1} $$

   So in our case:

   $$ \frac{ x^m - (x + h)^m }{h} = - \left( \binom{m}{1}x^{m - 1} + \binom{m}{2}x^{m - 2}h \enspace + ... + \enspace \binom{m}{m- 1}xh^{m - 2 } + h^{m - 1} \right) $$

   Now, our expression becomes:

   $$ (x^n)' = \lim_{h \to 0 } \enspace - \frac{\binom{m}{1}x^{m - 1} + \binom{m}{2}x^{m - 2}h \enspace + ... + \enspace \binom{m}{m- 1}xh^{m - 2 } + h^{m - 1} }{(x + h)^{m}x^{m}} $$

   When we pass to the limit when \( h \to 0 \) :

   $$ (x^n)' = - \frac{mx^{m - 1} }{x^{2m}} $$
   $$ (x^n)' = - mx^{-m - 1} $$

   However, as by hypothesis \(n = -m \),

   $$ \forall x \in \mathbb{R^*}, \enspace \forall n \in \Bigl[\mathbb{Z} \hspace{0.2em} \backslash \hspace{0.2em} \bigl \{ \mathbb{N} \bigr \}\Bigr], $$
   $$ (x^n)' = nx^{n - 1} $$

3. With an exponent \( n \) in the reals set \( \bigl( n \in\mathbb{R} \bigr) \)

   For an real exponent, this function is only defined for positive numbers \( (x \geqslant 0)\).

   So, considering any number \( X \neq 0 \) as \( e^{\ln\left(X\right)} \), we do have:

   $$ \forall x \in \mathbb{R^*}, \enspace \forall n \in \bigl[\mathbb{R} \hspace{0.2em} \backslash \mathbb{Q} \bigr],$$
   $$ x^n = e^{\ln(x^n)} $$

   And consequently,

   $$ (x^n)' = \left(e^{\ln(x^n)}\right)' $$

   With the properties of the natural logarithm , we know that:

   $$ \forall x \in \hspace{0.04em} \mathbb{R^*}, \ \forall n \in \hspace{0.04em} \mathbb{R}, $$

   $$\ln(x^n) = n . \ln(x)$$

   So,

   $$ (x^n)' = \left(e^{n.\ln(x)} \right)' $$

   We can then calculate this derivative by a chain derivation .

   It is known that:

   $$ \left(e^y \right)' = y'e^y $$

   So in our case:

   $$ (x^n)' = \bigl(n.\ln(x)\bigr)' e^{n .\ln(x)} $$
   $$ (x^n)' = \frac{n}{x} e^{n .\ln(x)} $$

   By reusing the previous property \( (2) \) in reverse:

   $$ (x^n)' = \frac{n}{x} e^{\ln(x^n)} $$
   $$ (x^n)' = nx^{-1} x^{n} $$
   $$ (x^n)' = nx^{n-1} $$

   And finally,

   $$ \forall x \in \hspace{0.04em} \mathbb{R}^*_+, \enspace \forall n \in \bigl[\mathbb{R} \hspace{0.2em} \backslash \mathbb{Q} \bigr], $$
   $$ (x^n)' = nx^{n - 1} $$

   Note: The case of rational exponents on \(\mathbb{R}^*_-\)

   The general proof above restricts the study to \(\mathbb{R}_+\) (strictly positive real numbers) because the analytical definition \(x^n = e^{n\ln(x)}\) requires the use of the natural logarithm.

   However, from a purely algebraic standpoint, there is a remarkable exception when the exponent \(n\) is a rational number with an odd denominator. We can then write:

   $$ n = \frac{p}{q} \hspace{3em} (p \in \mathbb{Z}, \ q \in 2\mathbb{N}+1) \text{ and } (p \land q = 1) $$

   In this specific case, the function \(f(x) = x^{\frac{p}{q}}\) can be defined on all of \(\mathbb{R}\) (or \(\mathbb{R}^*\) if \(p < 0\)) via the \(q\)-th root: \(x^{\frac{p}{q}}=\left(\sqrt[q]{x}\right)^p\). For example, the cube root function (\(x^{\frac{1}{3}}=\sqrt[3]{x}\)) perfectly accepts negative values (\(\sqrt[3]{-8}=-2\)).

   What happens to the derivative for \(x < 0\) ?

   The formula remains strictly the same: \((x^n)' = nx^{n-1}\). To prove this cleanly without the burden of absolute values, we use the parity of the function:

   • If \(p\) is even, the function is even (\(f(-x) = f(x)\)).
   • If \(p\) is odd, the function is odd (\(f(-x) = -f(x)\)).

   
   

   By introducing a change of variable \(X = -x\) (where \(X > 0\)), the chain rule allows us to instantly recover the general formula on \(]-\infty; 0[\).

   Case 1: \(p\) is even

   Since \(p\) is even, the function \(f\) is even, which means that \(f(x) = f(-x)\). To differentiate \(f\) on \(]-\infty; 0[\), we can therefore write:

   $$ f'(x) = \big(f(-x)\big)' $$

   Setting \(X = -x\) (with \(X > 0\)), we apply the chain rule :

   $$ f'(x) = -1 \cdot f'(X) $$

   Since \(X > 0\), the main proof applies: \(f'(X) = nX^{n-1}\). Replacing \(X\) with \(-x\), we obtain:

   $$ f'(x) = -n(-x)^{n-1} $$

   Let us determine the sign of the exponent: \(n-1 = \frac{p}{q} - 1 = \frac{p-q}{q}\). Since \(p\) is even and \(q\) is odd, their difference \((p-q)\) is an odd_integer. Consequently, the minus sign "comes out" of the power: \((-x)^{n-1} = -x^{n-1}\).

   Substituting this back into our equation, the two minus signs cancel out:

   $$ f'(x) = -n \cdot \big(-x^{n-1}\big) = nx^{n-1} $$

   Case 2: \(p\) is odd

   Since \(p\) is odd and \(q\) is odd, the function \(f\) is odd, which means that \(f(-x) = -f(x)\), or equivalently: \(f(x) = -f(-x)\). To differentiate \(f\) on \(]-\infty; 0[\), we write:

   $$ f'(x) = \big(-f(-x)\big)' = -\big(f(-x)\big)' $$

   Setting the change of variable \(X = -x\) (with \(X > 0\)), we apply the chain rule , which factors out a \(-1\):

   $$ f'(x) = -1 \cdot \big(-1 \cdot f'(X)\big) = f'(X) $$

   Since \(X > 0\), we can directly apply the formula proven in the first part of the course (\(f'(X) = nX^{n-1}\)). Replacing \(X\) with \(-x\), we obtain:

   $$ f'(x) = n(-x)^{n-1} $$

   Let us now determine the sign of the exponent: \(n-1 = \frac{p}{q} - 1 = \frac{p-q}{q}\). Since \(p\) is an odd integer and \(q\) is an odd integer, their difference \((p-q)\) is an even integer. Consequently, the effect of the minus sign disappears under the power: \((-x)^{n-1} = x^{n-1}\).

   Thus, we fall straight back onto the general formula:

   $$ f'(x) = nx^{n-1} $$

   
   

   Watch out for the pitfall: This extension to negative numbers is only stable if the fraction \(\frac{p}{q}\) is strictly irreducible . For example, \((-8)^{\frac{1}{3}} = -2\), but if we write the fraction in the equivalent form \(\frac{2}{6}\), the calculation \((-8)^{\frac{2}{6}} = \sqrt[6]{(-8)^2} = +2\) changes the sign of the result. It is to avoid these algebraic instabilities that in higher analysis, we prefer by convention to restrict non-integer powers to the interval \(]0; +\infty[\).

4. Conclusion

   As in the case of \(x\) is worth zero, we do have that:

   $$ \forall n \in \hspace{0.04em} \mathbb{R}^*, \ (x^n)' = 0$$

   Thus,

   $$\text{when } x \text{ is defined}, \ \forall n \in \mathbb{R}, $$
   $$ (x^n)' = nx^{n - 1} $$

The powers of n functions\(: (n^x)'\)

In this part, there will be many cases where \(n\) will be found under a logarithm , then for simplicity we will remove the case where \((n = 0)\).

Then, we specifically define a power of \(n\) function as followed:

With the definition of derivatives , we do have:

We are stucked with this expression, because we cannot directly dismiss \( h\).

Let us rewrite \( (3) \) under another form.

Consequently,

We can then calculate this derivative by a chain derivation :

As a result we do have,

The square root function\(: (\sqrt{x})' \)

The square root function is defined as followed:

1. By the definition of derivatives

   With the definition of derivatives , we do have:

   $$ \left(\sqrt{x} \right)' = \lim_{h \to 0 } \enspace \frac{\sqrt{x + h} - \sqrt{x}}{h} $$

   To rearrange this difference of two square roots, let us multiply up and down by the numerator's conjugate:

   $$ \left(\sqrt{x} \right)' = \lim_{h \to 0 } \enspace \frac{(\sqrt{x + h} - \sqrt{x})\textcolor{rgb(118 139 240)}{(\sqrt{x + h} + \sqrt{x})}}{h\textcolor{rgb(118 139 240)}{(\sqrt{x + h} + \sqrt{x})}} $$
   $$ \left(\sqrt{x} \right)' = \lim_{h \to 0 } \enspace \frac{x + h - x}{h(\sqrt{x + h} + \sqrt{x})} $$
   $$ \left(\sqrt{x} \right)' = \lim_{h \to 0 } \enspace \frac{h}{h(\sqrt{x + h} + \sqrt{x})} $$

   We can now simplify by \( h \):

   $$ \left(\sqrt{x} \right)' = \lim_{h \to 0 } \enspace \frac{1}{1(\sqrt{x + h} + \sqrt{x})} $$

   And finally,

   $$ \forall x \in \mathbb{R_+^*}, $$
   $$ \left(\sqrt{x} \right)' = \frac{1}{2\sqrt{x}} $$

2. Considerating the square root function as a power of x

   Considerating \( \sqrt{x} \) as a power of \( x \), we do have:

   $$ \sqrt{x} = x^{1 \over 2}$$

   From it, let us use the derivative of \( x^n\) :

   $$ \left(\sqrt{x} \right)' = (x^{1 \over 2})'$$
   $$ \left(\sqrt{x} \right)' = \frac{1}{2} x^{\frac{1}{2} - 1}$$
   $$ \left(\sqrt{x} \right)' = \frac{1}{2} x^{-\frac{1}{2}}$$
   $$ \left(\sqrt{x} \right)' = \frac{1}{2} \left(x^{\frac{1}{2}}\right)^{-1} = \frac{1}{2x^{\frac{1}{2}}}$$

   And finally,

   $$ \forall x \in \mathbb{R_+^*}, $$
   $$ \left(\sqrt{x} \right)' = \frac{1}{2\sqrt{x}} $$

   We can as well calculate any \(n^{th}\) root, with \(n \in \mathbb{Q}\).

3. Examples

   1. Calculation of \( \left(\sqrt[3]{x} \right)' \)

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

   2. Calculation of \( \left( x^{ \frac{5}{2} } \right)' \)

      $$ \left( x^{ \frac{5}{2} } \right)' =\frac{5}{2} x^{\frac{5}{2} - 1} $$
      $$ \left( x^{ \frac{5}{2} } \right)' =\frac{5}{2} x^{\frac{3}{2} } $$
      $$ \left( x^{ \frac{5}{2} } \right)' =\frac{5}{2} \sqrt{x^{3} } $$

The inverse function \( : \hspace{0.03em} \left(\frac{1}{x}\right)' \)

The inverse function is defined as followed:

Considerating \( \frac{1}{x} \) as a power of \( x \), we can use the derivative of \( x^n\) :

And finally,

The Napierian logarithm function\(: \Bigl[ \ln(x) \Bigr]'\)

1. The \(\ln(x)\) function

   The the Napierian logarithm function \((\ln(x))\) is defined as the reciprocal function of the exponential function \((e^x)\) :

   $$ \forall x \in \mathbb{R_+^*}, \enspace f(x) = \ln(x) = (e^x)^{-1}$$

   It can be defined by an integral :

   $$ \forall x \in \mathbb{R_+^*}, \enspace \ln(x) = \int^x_1 \frac{dt}{t}$$

   So,

   $$ \forall x \in \mathbb{R_+^*}, $$
   $$ \Bigl[ \ln(x) \Bigr]' = \frac{1}{x} $$

2. The \(\ln|x|\) function

   The Napierian logarithm function \((\ln(x))\) in absolute value is defined as:

   $$ \forall x \in \mathbb{R^*}, \enspace f(x) = \ln|x| = \Biggl \{ \begin{gather*} \forall x \in \mathbb{R_-^*}, \ f(x) = \ln(-x) \\ \forall x \in \mathbb{R_+^*}, \ f(x) = \ln(x) \end{gather*} $$

   We already have calculated the positive part above, so we need to calculate the negative one.

   Let us use the derivative of a reciprocal function :

   $$ \ln(-x)' = (-x)' \times \ln(-x)' $$
   $$ \ln(-x)' = - \frac{1}{-x} = \frac{1}{x}$$

   And finally,

   $$ \forall x \in \mathbb{R^*}, $$
   $$ \Bigl[ \ \ln|x| \ \Bigr]' = \frac{1}{x} $$

The logarithm to the base \(n\) function\(: \Bigl[ \log_n(x) \Bigr] '\)

The logarithm to the base \(n\) function is defined as followed:

Let us proceed as above with the neperian logarithm because the same logic is used.

Let \( g(x) = n^x \) be a function and \( g^{-1}(x) = \log_n(x) \) its reciprocal function.

And so,

We then find again the derivative of the neperian logarithm , also called natural logarithm.

Indeed, it is a specific case of the function \( \log_n(x) \) to the base \( e\).

The exponential function\(: (e^x)'\)

The exponential function is defined as the the derivative of a reciprocal function of the Napierian logarithm function \((\ln(x))\) :

By using the derivative of a reciprocal function , we do have:

Recap table of the standard functions derivatives