Remember our grand plan for calculating difficult
derivatives?
It had two ingredients: knowing the derivatives
of standard functions,
and knowing some rules of calculation.
In these videos we will tackle the derivatives
of some standard functions.
We will use the derivative of the exponential
function to calculate the derivatives of x^p.
The trick we will use here applies to any
ugly function involving exponentials,
so you can use it too!
Let us recall the definition of the derivative
as a limit of a difference quotient.
While the idea is not to use this formula
in actual calculations,
and use standard derivatives and rules of
calculation instead,
we have to use it to calculate the derivative
of our standard functions.
Let us first consider exponential functions.
We plug an exponential function f(x) = a^x in
the formula for the difference quotient.
This gives (a^(x + Δx) - a^x)/Δx,
but the rules of calculation for an exponential
shows that this also equals (a^x * a^(Δx) - a^x)/Δx.
Taking a^x apart we find that this equals
a^x * (a^(Δx) - 1)/Δx.
Notice that the right part is exactly the
difference quotient for a^x at 0.
Taking the limit as Δx goes to 0 we
see that the derivative of a^x at x equals
a^x times the derivative at 0,
a constant times a^x.
This constant depends on a.
Looking at the graphs of different exponential
functions,
we see that the derivative at zero is very
small and positive if a is slightly more than 1,
whereas we can make it as large as we want
by increasing a.
At some point the constant will be equal to
1,
so the graph will be tangent to y=x+1.
The corresponding value of a is called e.
The value of e equals 2.718 etc.
So in fact,
the derivative of the standard function e^x
is e^x by definition.
For a^x for other values of a we now use that
a^x = e^(x* ln(a)) and use the chain rule.
This gives that it is equal to the derivative of e^x at 
x*ln(a) times the derivative of x*ln(a)
which is equal to a^x * ln(a).
Let us now consider the derivative of ln(x).
ln(x) is the inverse function of e^x and we
can obtain its derivative by considering the
equation x = e^(ln(x)).
The derivative of x equals 1,
thus that is the result of applying the chain
rule to the right hand side.
So 1 equals the derivative of e^(ln(x)) 
equals [the derivative of e^x] at ln(x)
times the derivative of ln(x).
As e^(ln(x)) = x this gives 1 equals x times the derivative of ln(x).
So that find that the derivative of ln(x) equals 1/x.
The trick of using the exponential function
to calculate derivatives of a function involving
exponents can also be used if the x-dependence
is in the base.
Thus we find that the derivative of x^p (for
arbitrary p, not just integers!),
can be calculated by writing
x^p = e^(p*ln(x)).
Using the chain rule we obtain that the derivative
of x^p equals the derivative of e^x at p*ln(x)
times the derivative of p*ln(x),
which equals e^(p*ln(x)) times p/x,
which is x^p times p/x equals p times x^(p-1).
These calculations show that you don't have
to remember all standard derivatives as you
can derive some from others.
However,
we use the derivative of x^p often enough
that it is easier to just remember that it
is multiply by exponent and decrease exponent by 1,
than to remember its derivation.
However,
if you're like me and often forget whether
the derivative of 2^x is 2^x times ln(2) or
2^x divided by ln(2),
you can always fall back to remembering this
derivation to get it right.
Now try to use the trick to rewrite complicated
exponentials as e^something to calculate some
derivatives yourself.
