The derivatives of standard functions

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

A constant function is defined as followed:

Its derivative is:

$$ (\lambda)' = 0 $$

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

An affine function is defined as followed:

Its derivative is:

$$ (ax+ b)' = a $$

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

The absolute value function is defined as followed:

Its derivative is:

$$ |x|' = \frac{x}{|x|}$$

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

The square function is defined as followed:

Its derivative is:

$$ (x^2)' = 2x $$

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:

Its derivative is:

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

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:

Its derivative is:

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

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

The square root function is defined as followed:

Its derivative is:

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

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

The inverse function is defined as followed:

Its derivative is:

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

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:

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


$$ \forall x \in \mathbb{R_+^*}, $$

$$ \Bigl[ ln(x) \Bigr]' = \frac{1}{x} $$



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

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

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



$$ \forall x \in \mathbb{R^*}, $$

$$ \Bigl[ \ ln|x| \ \Bigr]' = \frac{1}{x} $$

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

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

Its derivative is:

$$ \Bigl[ log_n(x) \Bigr] ' = \frac{1}{x.ln(n)} $$

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

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

Its derivative is itself:

$$ (e^x)' = e^x$$

Recap table of the standard functions derivatives

Click on the title to access to the recap table.

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Demonstrations

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

A constant function is defined as followed:

With the definition of derivatives, we do have:

So,

$$ (\lambda)' = 0 $$

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,

$$ (ax + b)' = a $$

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

The absolute value function is defined as followed:

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

So,

$$ |x|' = \frac{x}{|x|}$$

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

The square function is defined as followed:

With the definition of derivatives, we do have:

So,

$$ (x^2)' = 2x $$

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:

$$\forall n \in \mathbb{N}, \enspace \forall (a, b) \in \hspace{0.05em} \mathbb{R}^2,$$

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

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

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

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

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 \( \mathbb{Z} \hspace{0.2em} \backslash \bigl\{ \mathbb{N} \bigr\} \), 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 \(n = -m \),

$$ \forall x \in \mathbb{R^*}, \enspace \forall n \in \Bigl[\mathbb{Z} \hspace{0.2em} \backslash \bigl\{ \mathbb{N} \bigr\}\Bigr], $$

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




3. With an exponent \( n \) in the rationnals set \( (n \in \mathbb{Q}) \)

A rational number \( n \) is written in the form:

$$ n = \frac{p}{q} \hspace{3em} (p \in \mathbb{Z}, \ q \in \hspace{0.05em} \mathbb{N}^*) $$

So, the number \( x^n \) can be written:

$$ x^n = x^{\frac{p}{q}} $$

As we go in the demonstrations, we will consider \( X \) as \( e^{ln\left(X\right)} \), we have to take into account the sign of the interior of the logarithm. And since a logarithm function does not accept any negative value, we must ensure that the number under it is always positive.

In the different cases that we are going to study, this will depend on the simplified form of the power \( \frac{p}{q} \).



1. If \( p \) is even:

So, the number \( p \) can be written:

$$ \forall a \in \mathbb{N}, \ p = 2a $$

Then,

$$ x^{\frac{p}{q}} = \left( x^{\frac{1}{q}} \right)^p $$
$$ x^{\frac{p}{q}} = \left( x^{\frac{1}{q}} \right)^{2a} $$
$$ x^{\frac{p}{q}} = \left( x^{\frac{a}{q}} \right)^{2} $$

The number \( x^{\frac{p}{q}} \) will be positive because it is a square. It can then be placed in a logarithm.



2. If \( q \) is even:

So, the number \( q \) can be written:

$$ \forall a \in \hspace{0.05em} \mathbb{N}^*, \ q = 2a $$

Then,

$$ x^{\frac{p}{q}} = \left( x^{p} \right)^{\frac{1}{q} }$$
$$ x^{\frac{p}{q}} = \left( x^{p} \right)^{\frac{1}{2a} }$$
$$ x^{\frac{p}{q}} = \left( x^{\frac{p}{a}} \right)^{\frac{1}{2} } $$

The number \( x^{\frac{p}{q}} \) will be positive and only defined for \( x \in \hspace{0.05em} \mathbb{R}^+ \) because it is a square root. It can then be placed in a logarithm.



3. If \( p \) and \( q \) are both odd:

The sign of \(x\) will be preserved after performing to power.

We will have to manage this case separately, because in the case of a number \( x < 0 \), the number \( x^{\frac{p}{q}} \) cannot be placed in a logarithm.



1. If \(p\) is even or \(q\) is even: \( \Bigl[ p \in \hspace{0.05em} \bigl \{ 2a \ | \ a \in \mathbb{Z} \bigr \}\Bigr] \lor \Bigl[ q \in \hspace{0.05em} \bigl \{ 2a \ | \ a \in \mathbb{N} \bigr \} \Bigr] \)

As we have seen above that in these two case, regardless of the sign of \(x\) the number \( x^{\frac{p}{q}} \) will always be positive, we will be able to place it under a logarithm.

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

$$ \begin{align*} \forall x \in \hspace{0.05em} \mathbb{R}^*, \enspace \forall p \in \hspace{0.05em} \Bigl \{ 2a \ | \ a \in \mathbb{Z} \Bigr \}, \ \forall q \in \mathbb{N}, \\ \hspace{8em} \underline{or} \\ \forall x \in \hspace{0.05em} \mathbb{R}^*_+, \enspace \forall p \in \mathbb{Z}, \ \forall q \in \hspace{0.05em} \Bigl \{ 2a \ | \ a \in \mathbb{N} \Bigr \}, \end{align*} \Biggr \} \hspace{1em} x^n = x^{\frac{p}{q}} = e^{ln\left(x^{\frac{p}{q}}\right)}$$

We then start from,

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

And consequently,

With the property \( (2) \) of the Napierian logarithm, we know that:

$$ \forall x \in \hspace{0.05em} \mathbb{R^*}, \ \forall n \in \hspace{0.05em} \mathbb{R}, $$

$$ln(x^n) = n . ln(x) \qquad (2) $$

So,

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

In the expression above, it was important to pass in absolute value because otherwise the study is restricted to \(x \in \hspace{0.05em} \mathbb{R}^*_+\).

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:

$$ (x^n)' = \Bigl(\frac{p}{q}.ln|x|\Bigr)' e^{\frac{p}{q} .ln|x|} $$
$$ (x^n)' = \frac{p}{qx} e^{\frac{p}{q} .ln|x|} $$

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

$$ (x^n)' = \frac{p}{qx} e^{ln\left(x^{\frac{p}{q}}\right)} $$
$$ (x^n)' = \frac{p}{q} x^{-1} x^{\frac{p}{q}} $$



Finally, by reallocating to the fraction \( \frac{p}{q} \) its initial value \(n\):

$$ \begin{align*} \forall x \in \hspace{0.05em} \mathbb{R}^*, \enspace \forall p \in \hspace{0.05em} \Bigl \{ 2a \ | \ a \in \mathbb{Z} \Bigr \}, \ \forall q \in \mathbb{N}, \\ \hspace{8em} \underline{or} \\ \forall x \in \hspace{0.05em} \mathbb{R}^*_+, \enspace \forall p \in \mathbb{Z}, \ \forall q \in \hspace{0.05em} \Bigl \{ 2a \ | \ a \in \mathbb{N} \Bigr \}, \end{align*}$$

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



2. If \( p \) and \( q \) are both odd: \( \Bigl[ p \in \hspace{0.05em} \bigl \{ 2a + 1 \ | \ a \in \mathbb{Z} \bigr \}\Bigr] \land \Bigl[ q \in \hspace{0.05em} \bigl \{ 2a + 1 \ | \ a \in \mathbb{N} \bigr \} \Bigr] \)

In this case, the sign of the number \( x \) is always preserved after performing the power.




To preserve a possible sign \( (-) \) after performing the power while using an absolute value under logarithm, we will use a sign generator in front of it to offset:

$$ \forall x \in \hspace{0.05em} \mathbb{R}^*, \enspace \forall p \in \hspace{0.05em} \Bigl \{ 2a + 1 \ | \ a \in \mathbb{Z} \Bigr \}, \ \forall q \in \hspace{0.05em} \Bigl \{ 2a + 1 \ | \ a \in \mathbb{N} \Bigr \}, $$

$$ x^n = x^{\frac{p}{q}} = \frac{x}{|x|} e^{ln \left |x^{\frac{p}{q}} \right|} \qquad (3) $$

And we then follow the same path as above:

$$ (x^n)' = \left( \frac{x}{|x|} e^{ \left( ln\left|x^{\frac{p}{q}} \right|\right)} \right)' $$
$$ (x^n)' = \left( \frac{x}{|x|} e^{\frac{p}{q}ln\left|x\right|} \right)' $$
$$ (x^n)' = \frac{x}{|x|} \Bigl(\frac{p}{q}.ln|x|\Bigr)' e^{\frac{p}{q} .ln|x|} $$
$$ (x^n)' = \frac{x}{|x|} \frac{p}{qx} e^{\frac{p}{q} .ln|x|} $$
$$ (x^n)' = \frac{x}{|x|} \frac{p}{qx} e^{ln \left |x^{\frac{p}{q}} \right|} $$

Here, we get our initial expression \((3) \) back:

$$ (x^n)' = \frac{p}{q} x^{-1} x^{\frac{p}{q}} $$



Finally, by reallocating to the fraction \( \frac{p}{q} \) its initial value \(n\):

$$ \forall x \in \hspace{0.05em} \mathbb{R}^*, \enspace \forall p \in \hspace{0.05em} \Bigl \{ 2a + 1 \ | \ a \in \mathbb{Z} \Bigr \}, \ \forall q \in \hspace{0.05em} \Bigl \{ 2a + 1 \ | \ a \in \mathbb{N} \Bigr \}, $$

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




4. With an exponent \( n \) in the irrationnals set \( \bigl( n \in\mathbb{R} \hspace{0.2em} \backslash \mathbb{Q} \bigr) \)

For an irrational 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:

$$ \forall x \in \hspace{0.05em} \mathbb{R^*}, \ \forall n \in \hspace{0.05em} \mathbb{R}, $$

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

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,

$$ \forall x \in \hspace{0.05em} \mathbb{R}^*_+, \enspace \forall n \in \bigl[\mathbb{R} \hspace{0.2em} \backslash \mathbb{Q} \bigr], $$

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




5. Conclusion

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

$$ \forall n \in \hspace{0.05em} \mathbb{R}^*, \ (x^n)' = 0$$

Thus,

$$when \ x \ is \ defined, \ \forall n \in \mathbb{R}, $$

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

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,

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

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:

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

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,

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

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:

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




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,

$$ \Bigl[ log_n(x) \Bigr] ' = \frac{1}{x.ln(n)} $$

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:

$$ (e^x)' = e^x$$

Recap table of the standard functions derivatives