Question

What is the derivative of $3{{e}^{\left( 2x+1 \right)}}$ ?

Answer 1

To differentiate a complicated function $f\left( g\left( x \right) \right)$ , we must use the Chain Rule of differentiation, which says
$\dfrac{d\left( f\left( g\left( x \right) \right) \right)}{dx}=\dfrac{d\left( f\left( g\left( x \right) \right) \right)}{d\left( g\left( x \right) \right)}\times \dfrac{d\left( g\left( x \right) \right)}{dx}$
We should also use the identities $\dfrac{d\left( {{e}^{x}} \right)}{dx}={{e}^{x}}$ and $\dfrac{d\left( af\left( x \right) \right)}{dx}=a\dfrac{df\left( x \right)}{dx}$ , to get the final result.

Complete step by step solution:
In our question, we need to find the derivative of $3{{e}^{\left( 2x+1 \right)}}$ .
Here, nothing is specified, so we should assume that differentiation is to be done with respect to x.
This implies that we need to find $\dfrac{d\left( 3{{e}^{\left( 2x+1 \right)}} \right)}{dx}$ .
We know that if a is a constant, then we have $\dfrac{d\left( af\left( x \right) \right)}{dx}=a\dfrac{df\left( x \right)}{dx}$.
So, by using this property, we can write
$\dfrac{d\left( 3{{e}^{\left( 2x+1 \right)}} \right)}{dx}=3\dfrac{d\left( {{e}^{\left( 2x+1 \right)}} \right)}{dx}...\left( i \right)$
We know that a complicated function like $f\left( g\left( x \right) \right)$ can be differentiated using the Chain Rule of Differentiation. According to the Chain Rule of Differentiation, we can write
$\dfrac{d\left( f\left( g\left( x \right) \right) \right)}{dx}=\dfrac{d\left( f\left( g\left( x \right) \right) \right)}{d\left( g\left( x \right) \right)}\times \dfrac{d\left( g\left( x \right) \right)}{dx}$
Thus, we can write
$\dfrac{d\left( {{e}^{\left( 2x+1 \right)}} \right)}{dx}=\dfrac{d\left( {{e}^{\left( 2x+1 \right)}} \right)}{d\left( 2x+1 \right)}\times \dfrac{d\left( 2x+1 \right)}{dx}...\left( ii \right)$
Now, let us calculate $\dfrac{d\left( {{e}^{\left( 2x+1 \right)}} \right)}{d\left( 2x+1 \right)}\text{ and }\dfrac{d\left( 2x+1 \right)}{dx}$ separately.
For this, we should assume $y=2x+1$ .
Thus, we have
$\dfrac{d\left( {{e}^{\left( 2x+1 \right)}} \right)}{d\left( 2x+1 \right)}=\dfrac{d\left( {{e}^{y}} \right)}{dy}...\left( iii \right)$
We know the identity $\dfrac{d\left( {{e}^{x}} \right)}{dx}={{e}^{x}}$
So, it is very clear that,
$\dfrac{d\left( {{e}^{y}} \right)}{dy}={{e}^{y}}$
Using this value into the RHS of equation (iii), we get
$\dfrac{d\left( {{e}^{\left( 2x+1 \right)}} \right)}{d\left( 2x+1 \right)}={{e}^{y}}$
Putting the value of $y=2x+1$ , we get
$\dfrac{d\left( {{e}^{\left( 2x+1 \right)}} \right)}{d\left( 2x+1 \right)}={{e}^{\left( 2x+1 \right)}}...\left( iv \right)$
Now, we will evaluate the second part of equation (ii), i.e., $\dfrac{d\left( 2x+1 \right)}{dx}$.
We know that it could also be expressed as
$\dfrac{d\left( 2x+1 \right)}{dx}=\dfrac{d\left( 2x \right)}{dx}+\dfrac{d\left( 1 \right)}{dx}$
Thus, we get $\dfrac{d\left( 2x+1 \right)}{dx}=2...\left( v \right)$
Substituting the values from equation (iv) and equation (v) into equation (ii), we get
$\dfrac{d\left( {{e}^{\left( 2x+1 \right)}} \right)}{dx}={{e}^{\left( 2x+1 \right)}}\times 2$
Thus, we have $\dfrac{d\left( {{e}^{\left( 2x+1 \right)}} \right)}{dx}=2{{e}^{\left( 2x+1 \right)}}$.
And, using equation (i), we can write
$\dfrac{d\left( 3{{e}^{\left( 2x+1 \right)}} \right)}{dx}=3\times 2{{e}^{\left( 2x+1 \right)}}$
$\Rightarrow \dfrac{d\left( 3{{e}^{\left( 2x+1 \right)}} \right)}{dx}=6{{e}^{\left( 2x+1 \right)}}$
Hence, the derivative of $3{{e}^{\left( 2x+1 \right)}}$ is $6{{e}^{\left( 2x+1 \right)}}$.

Note: We must remember that Chain Rule has a high probability of error. So, we must take utmost care while using this concept and must not skip any step in between.
We must also remember all formulae of derivatives by heart, as without them, we will not be able to solve the problems based on this concept.