Question

How do you find the 1st and 2nd derivative of $2{e^{2x}}$?

Answer 1

In order to find derivatives of an expression, we should know the basic rules of differentiation like constant multiple rule i.e., $\dfrac{d}{{dx}}\left( {cf(x)} \right) = cf'(x)$, chain rule i.e., $\dfrac{d}{{dx}}f(g(x)) = f'(g(x)) \times g'(x)$ and double differentiation rule i.e., $\dfrac{{{d^2}f(x)}}{{d{x^2}}} = \dfrac{d}{{dx}}\left( {\dfrac{{df(x)}}{{dx}}} \right)$. Here, we have to apply these three rules to find out the first and second derivative of the given expression.

Complete step by step solution:
(i)
We are given:
$2{e^{2x}}$
Let us assume the given expression as $y$:
$y = 2{e^{2x}}$
First, we need to find its 1st derivative i.e., $\dfrac{{dy}}{{dx}}$.
Therefore,
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {2{e^{2x}}} \right)$
As we know the constant multiple rule i.e.,
$\dfrac{d}{{dx}}\left( {cf(x)} \right) = cf'(x)$
Applying this rule on our expression, we get:
$\dfrac{{dy}}{{dx}} = 2\dfrac{d}{{dx}}\left( {{e^{2x}}} \right)$
(ii)
Now, as we have to differentiate ${e^{2x}}$, we will assume:
$f(x) = {e^x}$
And,
$g(x) = 2x$
Then, we can say that:
$f(g(x)) = f(2x) = {e^{2x}}$

Since, we know the chain rule i.e.,
$\dfrac{d}{{dx}}f(g(x)) = f'(g(x)) \times g'(x)$
Therefore, we get:
$2\dfrac{d}{{dx}}\left( {{e^{2x}}} \right) = 2\left( {{e^{2x}} \times 2} \right)$
Since, we know that
$\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}$ and $\dfrac{d}{{dx}}\left( {2x} \right) = 2$]
Simplifying it further,
$\dfrac{d}{{dx}}\left( {2{e^{2x}}} \right) = 4{e^{2x}}$
(iii)
Now, as we have calculated the 1st derivative of ${e^{2x}}$, in order to calculate its 2nd derivative, we need to differentiate its 1st derivative again.
$\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right)$ -[eq. 1]
As we know that,
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {2{e^{2x}}} \right) = 4{e^{2x}}$
Putting the value of $\dfrac{{dy}}{{dx}}$ in the eq. 1, we get:
$\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}\left( {4{e^{2x}}} \right)$
(iv)
Again, applying the constant multiple rule here,
$\dfrac{d}{{dx}}\left( {cf(x)} \right) = cf'(x)$
We get,
$\dfrac{{{d^2}y}}{{d{x^2}}} = 4\dfrac{d}{{dx}}\left( {{e^{2x}}} \right)$
(v)
Here, again we will apply the chain rule
$\dfrac{d}{{dx}}f(g(x)) = f'(g(x)) \times g'(x)$
We will obtain,
$\dfrac{{{d^2}y}}{{d{x^2}}} = 4\left( {{e^{2x}} \times 2} \right)$
On simplifying, we get:
$\dfrac{{{d^2}y}}{{d{x^2}}} = 8{e^{2x}}$
Hence, the 1st derivative of $2{e^{2x}}$ is $4{e^{2x}}$ and the 2nd derivative is $8{e^{2x}}$.

Note: Whenever we need to find the derivative of the function ${e^{f(x)}}$ where $f(x)$ is a function of $x$, we can directly write the function itself multiplied by the derivative of its power as the derivative of ${e^x}$ is ${e^x}$ itself and if we have to find further derivatives, we just have to multiply its power’s derivative with the previous derivative.