Calculus/Derivatives of Exponential and Logarithm Functions

We shall first look at the irrational number ${\displaystyle e}$ in order to show its special properties when used with derivatives of exponential and logarithm functions. As mentioned before in the Algebra section, the value of ${\displaystyle e}$ is approximately ${\displaystyle e\approx 2.718282}$ but it may also be calculated as the Infinite Limit:

${\displaystyle e=\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}}$

Now we find the derivative of ${\displaystyle \ln(x)}$ using the formal definition of the derivative:

${\displaystyle {\frac {d}{dx}}\ln(x)=\lim _{h\to 0}{\frac {\ln(x+h)-\ln(x)}{h}}=\lim _{h\to 0}{\frac {1}{h}}\ln \left({\frac {x+h}{x}}\right)=\lim _{h\to 0}\ln \left({\frac {x+h}{x}}\right)^{\frac {1}{h}}}$

Let ${\displaystyle n={\frac {x}{h}}}$ . Note that as ${\displaystyle n\to \infty }$ , we get ${\displaystyle h\to 0}$ . So we can redefine our limit as:

${\displaystyle \lim _{n\to \infty }\ln \left(1+{\frac {1}{n}}\right)^{\frac {n}{x}}={\frac {1}{x}}\ln \left(\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}\right)={\frac {1}{x}}\ln(e)={\frac {1}{x}}}$

Here we could take the natural logarithm outside the limit because it doesn't have anything to do with the limit (we could have chosen not to). We then substituted the value of ${\displaystyle e}$ .

If we wanted, we could go through that same process again for a generalized base, but it is easier just to use properties of logs and realize that:

${\displaystyle \log _{a}(x)={\frac {\ln(x)}{\ln(a)}}}$

Since ${\displaystyle {\frac {1}{\ln(a)}}}$ is a constant, we can just take it outside of the derivative:

${\displaystyle {\frac {d}{dx}}\log _{a}(x)={\frac {1}{\ln(a)}}\cdot {\frac {d}{dx}}\ln(x)}$

Which leaves us with the generalized form of:

An alternative approach to derivative of the logarithm refers to the original expression of the logarithm as quadrature of the hyperbola y = 1/x . This approach is described in an extension of precalculus in § 1.8.

We shall take two different approaches to finding the derivative of ${\displaystyle a^{x}}$ . The first by using the limit definition and the Squeeze Theorem. The second approach will make use of the fact that ${\displaystyle \log _{a}(a^{x})=x}$. The first approach follows.

Proof

Let ${\displaystyle f(x)=a^{x}}$, then:

${\displaystyle f'(x)=\lim _{h\to 0}{\frac {a^{x+h}-a^{x}}{h}}=a^{x}\lim _{h\to 0}{\frac {a^{h}-1}{h}}}$

Making the limit substitution: ${\displaystyle h={\frac {t}{\ln(a)}}\implies a^{h}=a^{\frac {t}{\ln(a)}}=e^{t}}$ and noting that as ${\displaystyle h\to 0,h\ln(a)=t\to 0}$ we have:

${\displaystyle a^{x}\lim _{t\to 0}{\frac {e^{t}-1}{\frac {t}{\ln(a)}}}=a^{x}\ln(a)\lim _{t\to 0}{\frac {e^{t}-1}{t}}}$

From the limit ${\displaystyle e^{t}=\lim _{n\to \infty }\left(1+{\frac {t}{n}}\right)^{n}}$ we have,

${\displaystyle e^{t}=\lim _{n\to \infty }\left(1+{\frac {t}{n}}\right)^{n}=\lim _{n\to \infty }\sum _{k=0}^{n}{\binom {n}{k}}\left({\frac {t}{n}}\right)^{k}=1+t+O(t^{2})\geq 1+t}$

From this inequality one has

${\displaystyle e^{t}\geq t+1\implies {\begin{cases}{\frac {e^{t}-1}{t}}\geq 1,&t>0\\1\geq {\frac {e^{t}-1}{t}},&t<0\end{cases}}}$

One also has,

${\displaystyle e^{t}\geq 1+t\implies e^{-t}\leq 1-t\implies e^{t}\leq {\frac {1}{1-t}}}$

Similarly,

${\displaystyle e^{t}\leq {\frac {1}{1-t}}\implies {\begin{cases}{\frac {e^{t}-1}{t}}\leq {\frac {{\frac {1}{1-t}}-1}{t}},&t>0\\{\frac {e^{t}-1}{t}}\geq {\frac {{\frac {1}{1-t}}-1}{t}},&t<0\end{cases}}={\begin{cases}{\frac {e^{t}-1}{t}}\leq {\frac {1}{1-t}},&t>0\\{\frac {e^{t}-1}{t}}\geq {\frac {1}{1-t}},&t<0\end{cases}}}$

Squeezing the left-hand limit, when ${\displaystyle t<0}$:

${\displaystyle {\frac {1}{1-t}}\leq {\frac {e^{t}-1}{t}}\leq 1\implies \lim _{t\to 0^{-}}{\frac {1}{1-t}}\leq \lim _{t\to 0^{-}}{\frac {e^{t}-1}{t}}\leq \lim _{t\to 0^{-}}1}$

Since ${\displaystyle \lim _{t\to 0^{-}}{\frac {1}{1-t}}=1}$ by Squeeze Theorem:

${\displaystyle L^{-}=\lim _{t\to 0^{-}}{\frac {e^{t}-1}{t}}=1}$

The same exercise may be repeated for the right-hand limit to get ${\displaystyle L^{+}=1}$. Since both ${\displaystyle L^{-}=L^{+}=1}$, then:

${\displaystyle L=\lim _{t\to 0}{\frac {e^{t}-1}{t}}=1}$

Hence the whole derivative becomes:

${\displaystyle f'(x)=a^{x}\ln(a)\lim _{t\to 0}{\frac {e^{t}-1}{t}}=a^{x}\ln(a)}$ □

As desired. We begin with the second proof:

Proof

We know that ${\displaystyle \log _{a}(a^{x})=x}$. Taking the derivative of both sides:

${\displaystyle {\frac {d}{dx}}[\log _{a}(a^{x})]={\frac {d}{dx}}[x]\implies {\frac {1}{\ln(a)a^{x}}}{\frac {d}{dx}}[a^{x}]=1}$

This expression comes from the chain rule and log rule derived from before. Now rearranging:

${\displaystyle {\frac {d}{dx}}[a^{x}]=a^{x}\ln(a)}$ □

As desired. The base for which the derivative is itself is ${\displaystyle a=e}$. A proof of this lemma follows.

Proof

We know the derivative of ${\displaystyle a^{x}}$ is ${\displaystyle a^{x}\ln(a)}$. One may use the substitution ${\displaystyle a=e}$ noting that ${\displaystyle \ln(e)=1}$. Then we have:

${\displaystyle {\frac {d}{dx}}[e^{x}]=e^{x}}$ □

As desired.

We can use the properties of the logarithm, particularly the natural log, to differentiate more difficult functions, such a products with many terms, quotients of composed functions, or functions with variable or function exponents. We do this by taking the natural logarithm of both sides, re-arranging terms using the logarithm laws below, and then differentiating both sides implicitly, before multiplying through by ${\displaystyle y}$ .

See the examples below.

We shall now prove the validity of the power rule using logarithmic differentiation.

${\displaystyle {\frac {d}{dx}}\ln(x^{n})=n{\frac {d}{dx}}\ln(x)=nx^{-1}}$

${\displaystyle {\frac {d}{dx}}\ln(x^{n})={\frac {1}{x^{n}}}\cdot {\frac {d}{dx}}x^{n}}$

Thus:

${\displaystyle {\frac {1}{x^{n}}}\cdot {\frac {d}{dx}}x^{n}=nx^{-1}}$

${\displaystyle {\frac {d}{dx}}x^{n}=nx^{n-1}}$

Example 2

Example 3

Example 4