### 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:

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

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}#.

**Pass Your Math**independent of your university. See pricing and more.

Or visit omptest.org if jou are taking an OMPT exam.