Question

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

Answer 1

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 \dfrac{d}{{dx}}\left\\{ {f\left( {g\left( x \right)} \right)} \right\\} = \dfrac{d}{{dt}}\left( {{e^t}} \right) \times \dfrac{d}{{dx}}\left( {2x} \right)
Simplifying the above equation, we will then obtain the following equation with us:-
\Rightarrow \dfrac{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 \dfrac{d}{{dx}}\left\\{ {f\left( {g\left( x \right)} \right)} \right\\} = 2{e^t}
Replacing t back by 2x, we will then obtain the following equation with us:-
\Rightarrow \dfrac{d}{{dx}}\left\\{ {f\left( {g\left( x \right)} \right)} \right\\} = 2{e^{2x}}

Hence, $\dfrac{d}{{dx}}\left( {{e^{2x}}} \right) = 2{e^{2x}}$ 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 \dfrac{d}{{dx}}\left\\{ {f\left( {g(x)} \right)} \right\\} = f'(g(x)) \times g'(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:-
$\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}$
This can be termed as: Differentiation of an exponential function does not change at all, it remains the same always.
$\dfrac{d}{{dx}}\left( {c{x^n}} \right) = c\dfrac{d}{{dx}}\left( {{x^n}} \right) = c(n - 1){x^{n - 1}}$, 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:-
$\Rightarrow \dfrac{d}{{dx}}\left( {c.f(x)} \right) = c\dfrac{d}{{dx}}\left( {f(x)} \right)$, where c is the constant.