VIDEO: Perform a derivative a to the power of x

That is a derivation

Derivation is a term from the mathematics, more precisely from differential calculus.

• The derivative of a function at a point x indicates the slope of the function at precisely this point.

• The following notations are used for the derivation in mathematics: f ^'(x) or df (x) / dx.

• For this reason, the differential calculus, including the derivation of Functions, basically with the Curve discussion used.

Also in the field of physics deliver Derivatives important findings. So one can deduce the instantaneous velocity of a particle by deriving the position-time function.

How to differentiate a function "a to the power of x"

Like everything else in mathematics, differential calculus is subject to strict rules. So it is up to you to decide anew for each function which rules and procedures you will use. To derive the function "a to the power of x", just proceed as follows:

1. First, write down the task. In this case, the following applies in the case of "a to the power of x": f (x) = a^x, wanted is f ^'(x) or df (x) / dx. Since rules such as the chain rule do not work for such functions, you must first transform this function to be "derivative-friendly". You can do this by a^x bring into the Euler representation. The function e^x can be derived easily.
2. The logarithm naturalis helps us with the transformation. This provides us with the following display options: a^b = e^b* ln (a). So you can represent f (x) as follows: f (x) = a^x = e^{x * ln (a)}. You can now easily derive this function.
3. Use the chain rule here. This says: f ^'(u (x)) = f ^'(u (x)) * u^'(x). To do this, substitute u (x) for v. In this case v = x * ln (a).
4. This results in the following new notation for our chain rule: f ^'(v) = f ^'(v) * v^'.
5. In the event of e^{x * ln (a)} the result is: f ^'(v) = (e^v)^' * v^'. Now you can easily derive the individual terms.
6. e^v always stays e^v.
7. v^' = (x * ln (a))^' = ln (a), since the derived x results in 1 and pre-factors remain.
8. So after back substitution of v we get the following: f ^'(x) = (a^x)^' = (e^{x * ln (a)} )^' = e^{x * ln (a)} * ln (a).

With a^x = e^{x * ln (a)} so we come to the end result: (a^x)^' = ax * ln (a).