I immerse myself in nuclear physics; where e occurs frequently. I know the derivative of e raised to the x power is the same, but how do you differentiate a function with the variable as the exponent?
Answer
So the question is, how do you lead the function
f(x) = aX
af, with a being a strict positive number (different from e, and from 1). So an exponential function with base a. The formula for this is f ‘ (x) = aX . ln a
This is very easy to find : since aX always strictly positive you can always take a natural logarithm and then the e-power again. Since the two are each other’s inverse, they cancel each other out. So :
f(x) = exp [Â ln ( ax) ]
So by exp I just mean “e to the power”.
Now you know that a logarithm has the following property : ln ab = b . ln a
So : f(x) = exp [ x ln(a) ]
The exponential function aX has thus now been converted into an “ordinary” exponential function.
ln(a) is simply a constant. So now just deduce ;
f'(x) = exp [ x ln(a) ] . ln(a) = exp [Â ln(ax) ] . ln(a) = aX . ln(a)
Note that if a is the number e, you indeed “just” have eX finds as a derivative of eX.
After all, ln(e) = 1
Another important consequence : the derivative of af(x)
is thus with the chain rule : af(x) . ln(a) . f’ (x)
Answered by
prof.dr. Paul Hellings
Department of Mathematics, Fac. IIW, KU Leuven
Old Market 13 3000 Leuven
https://www.kuleuven.be/
.