Exponential functions and logarithm derivatives revisited

Rules of differentiation: Rules of computation for the derivative

Exponential functions and logarithm derivatives revisited

Now that we have dealt with the chain rule, we look again at the derivatives of the power functions.

Exponential rule for differentiation

Let $a$ be a positive real number.

The derivative of the function $a^x$ is $\ln(a)\cdot a^x$ .

Since $a^x = \left({\e}^{\ln(a)}\right)^x = {\e}^{\ln(a)\cdot x} = \exp(\ln(a)\cdot x)$ , the chain rule gives $\begin{array}{rcl} \dfrac{\dd}{{\dd}x}\left(a^x\right) &=&\exp'(\ln(a)\cdot x)\cdot \dfrac{\dd}{{\dd}x}(\ln(a)\cdot x)\\ &=& \exp(\ln(a)x)\cdot \ln(a)\\&=&\exp(\ln(a^x))\cdot \ln(a)\\ &=& a^x \cdot \ln(a)\,\tiny.\end{array}$ This proves the theorem.

We can use the same rule to determine the derivative of the natural logarithm.

Logarithmic rule for differentiation

Let $a$ be a positive real number distinct from $1$ .

The derivative of the function $\log_a(x)$ is $\dfrac{1}{\ln(a)\cdot x}$ .

In particular, the derivative of the function $\ln(x)$ is $\dfrac{1}{x}$ .

First consider $f(x)=\ln(x)$ . Then $\ee^{f(x)}=x$ . If we take the derivative on both sides of the equality and apply the chain rule, we obtain the following equality:

$\ee^{f(x)} \cdot f'(x)=1\tiny.$

Since $\ee^{f(x)}=x$ , this gives $x\cdot f'(x)=1$ , so $f'(x)=\frac{1}{x}$ . This proves the special case.

The general case can be derived from it as follows:

$\begin{array}{rcl}\dfrac{\dd}{\dd x}\left(\log_a(x)\right)&=&\dfrac{\dd}{\dd x}\left(\dfrac{\ln(x)}{\ln(a)}\right)\\ &=&\dfrac{1}{\ln(a)}\cdot\dfrac{\dd}{\dd x}\left(\ln(x)\right)\\ &=&\dfrac{1}{\ln(a)}\cdot\dfrac{1}{x}\\ &=&\dfrac{1}{\ln(a)\cdot x} \end{array}$

Determine the derivative of $f(t)= 2^t$ .

$f'(t)=$ $\ln \left(2\right)\cdot 2^{t}$

We apply the exponential rule for differentiation with $a=2$ : the derivative of the function $f(t)=2^t$ is $f'(t)=\ln \left(2\right)\cdot 2^{t}$ .

New example