In the
first section we looked at the
formula *A*^{B} as an algebraic object, when *A* and
*B* are fixed numbers. Then we imagined what would
happen if we tried to make *A* or *B* into a variable, thus
obtaining
powers as functions and
exponentials. What happens if we
have a "general power", that is, a power where a variable *x* is present
both in the base *A* and the exponent *B*?

The simplest example is *x*^{x},*x*)]^{2x + cos(x)}.

Unlike the previous two cases, general powers are hard to grasp based on the
properties of the usual power. In fact, if we tried this approach, looking at
general powers through the lens of algebraic powers, we would get very weird
things. One does not have to go far, when given a function, we always start
by determining its domain. What happens if we ask for which values of
*x* does the expression *x*^{x}

This shows that we have to adopt a wholly new approach to general powers. The previous section holds the key, we have a formula there that transforms general powers into products.

Definition.

Letf,gbe functions. We define thegeneral powerby f^{g}

f(x)^{g(x)}=e^{ln[f (x)]⋅g(x)}=e^{g(x)⋅ln[f (x)]}.

For instance, the meaning of the function
*x*^{x}*e*^{x⋅ln(x)}.

Note that in the expression on the right there are no general powers, just a
simple composition of logarithm, multiplication and exponential, all of which
are nice operations. Everything that we do with functions that are general
powers - domain, limits, derivative - must be done in this "*e* to
logarithm" form. What do we get?

By looking at the definition of
*f* ^{g},*e*^{ln[f (x)] g(x)}*f* (*x*)*g*(*x*)*f* (*x*)]*f* (*x*) > 0.

Thus we get the following:

*D*( *f* ^{g}) =
{*x*∈*D*( *f* ) ∩ *D*(*g*); *f* (*x*) > 0}.

Similarly, differentiating *f* ^{g}*e*^{ln[f (x)] g(x)}.

*D*( *f* ^{g})
= *e*^{ln( f )g} ⋅ [ln( *f* )*g*]′ = *f* ^{g} ⋅ [(*f* ′/*f* )*g* + ln( *f* )*g*′].

Usually people do not remember these formulas, because they are not needed.
The only thing that is needed is this basic rule:
**When working with general powers, always rewrite them to the " e to
logarithm" form**.

**Example:** Find the domain and derivative of
*x*^{x}.

**Solution:** We have
*x*^{x} = *e*^{x⋅ln(x)}.

*D*(*x*^{x}) = (0,∞).

We also easily find the derivative:

*x*^{x}]′ =
[*e*^{xln(x)}]′ =
*e*^{xln(x)} ⋅ [*x*ln(*x*)]′ =
*x*^{x} ⋅ [ln(*x*) + *x*(1/*x*)] =
*x*^{x} ⋅ [ln(*x*) + 1].

For more information about this function see appropriate problem in Derivatives - Exercises - Graphing functions.

**Remark:** One mistake students often make is that they try to weasel out
of the (admittedly more complicated) "*e* to logarithm" way. Typically they
would try to pretend that the base or the exponent is just some number and
use appropriate formulas. Of course this is wrong, as we can see:

We pretend that *x*^{x} is a power and apply the
appropriate rule for derivative:

*x*^{x}]′ =
*x*⋅*x*^{x−1} =
*x*^{x}.

We pretend that *x*^{x} is a an exponential with base
*x* and apply the appropriate rule for derivative:

*x*^{x}]′ =
ln(*x*)⋅*x*^{x}.

If I got a penny every time I see one of these two incorrect calculations on a test, I would, like, have all those pennies. Take this as a warning and try to do it right.

Trigonometric functions

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functions