Question

How do I find the derivative of $f(x)=e^{2x}$?

Answer 1

Hint: We will use the chain rule in which we will take f (x) to be $e^x$ and g (x) to be 2x, then we will just find the required derivative as the answer.

Complete step-by-step solution:
We are given that we are required to find the derivative of $f(x)=e^{2x}$.
Let us assume that $f(x)=e^x$ and g (x) = 2x.
Now, we know that we can use the chain rule to find out the following expression by replacing 2x by t in f (x) in the right hand side:-
$\Rightarrow {\displaystyle \frac{d}{dx}}\left\{f\left(g\left(x\right)\right)\right\}={\displaystyle \frac{d}{dt}}\left(e^t\right)\times {\displaystyle \frac{d}{dx}}\left(2x\right)$
Simplifying the above equation, we will then obtain the following equation with us:-
$\Rightarrow {\displaystyle \frac{d}{dx}}\left\{f\left(g\left(x\right)\right)\right\}=e^t\times 2$
Simplifying the above equation further, we will then obtain the following equation with us:-
$\Rightarrow {\displaystyle \frac{d}{dx}}\left\{f\left(g\left(x\right)\right)\right\}=2e^t$
Replacing t back by 2x, we will then obtain the following equation with us:-
$\Rightarrow {\displaystyle \frac{d}{dx}}\left\{f\left(g\left(x\right)\right)\right\}=2e^{2x}$

Hence, ${\displaystyle \frac{d}{dx}}\left(e^{2x}\right)=2e^{2x}$ is the required answer.

Note: The students must note that the chain rule is given by the following equation:-
If we have two functions named as f (x) and g (x), then we have the following expression with us:-
$\Rightarrow {\displaystyle \frac{d}{dx}}\left\{f\left(g(x)\right)\right\}=f^{\prime }(g(x))\times g^{\prime }(x)$
The students must also note that we replaced 2x by t in the equation while differentiating because that made things easy for us to look at. Because we had to write f (g (x) ) in there, we just assumed that g (x) = 2x = t, therefore, it became f (t) instead of f (g (x) ).
The students must note the following rules and formulas for differentiation of any functions:-
${\displaystyle \frac{d}{dx}}\left(e^x\right)=e^x$
This can be termed as: Differentiation of an exponential function does not change at all, it remains the same always.
${\displaystyle \frac{d}{dx}}\left(c{x}^n\right)=c{\displaystyle \frac{d}{dx}}\left(x^n\right)=c(n-1)x^{n-1}$, where c is a constant.
We always can take the constant term out of the differentiation and sign and work on the rest function inside the brackets. This can be termed in mathematical words as follows:-
$\Rightarrow {\displaystyle \frac{d}{dx}}\left(c.f(x)\right)=c{\displaystyle \frac{d}{dx}}\left(f(x)\right)$, where c is the constant.