How do you find the derivative of \[{{e}^{-6x}}+e\]?

Hint: To solve the given question, we should know the derivatives of some of the functions, and how to differentiate composite functions. The composite functions are functions of the form \[f\left( g(x) \right)\], their derivative is found as, \[\dfrac{d\left( f\left( g(x) \right) \right)}{dx}=\dfrac{d\left( f\left( g(x) \right) \right)}{d\left( g(x) \right)}\dfrac{d\left( g(x) \right)}{dx}\]. The functions whose derivatives we should know are \[{{e}^{x}}\]\[,ax\] and constant function, their derivatives are \[{{e}^{x}},a\And 0\] respectively. We will use these to find the derivative of the given function. We should also know that the derivative can be separated over addition.

Complete step by step solution:
We know that the derivative of the composite function is evaluated as \[\dfrac{d\left( f\left( g(x) \right) \right)}{dx}=\dfrac{d\left( f\left( g(x) \right) \right)}{d\left( g(x) \right)}\dfrac{d\left( g(x) \right)}{dx}\].
We are given the function \[{{e}^{-6x}}+e\], we are asked to find its derivative. As we know that the derivative can be separated over addition of function. So, we can evaluate its derivative as,
\[\dfrac{d\left( {{e}^{-6x}}+e \right)}{dx}=\dfrac{d\left( {{e}^{-6x}} \right)}{dx}+\dfrac{d\left( e \right)}{dx}\]
The \[{{e}^{-6x}}\] is a composite function of the form \[f\left( g(x) \right)\], here we have \[f(x)={{e}^{x}}\And g(x)=-6x\].
To find the derivative of the given function, we need to find, \[\dfrac{d\left( {{e}^{-6x}} \right)}{d\left( -6x \right)}\And \dfrac{d\left( -6x \right)}{dx}\].
We know that the derivative of \[{{e}^{x}}\] with respect to x is \[{{e}^{x}}\] itself. Thus, the derivative of \[{{e}^{-6x}}\] with respect to \[-6x\] must be equal to \[{{e}^{-6x}}\]. Hence, we get \[\dfrac{d\left( {{e}^{-6x}} \right)}{d\left( -6x \right)}={{e}^{-6x}}\]. Also, the derivative of \[-6x\] with respect to x is 6.
The e is a constant function here, hence its derivative will be zero.
\[\dfrac{d\left( {{e}^{-6x}}+e \right)}{dx}=\dfrac{d\left( {{e}^{-6x}} \right)}{dx}+\dfrac{d\left( e \right)}{dx}\]
\[\Rightarrow \dfrac{d\left( {{e}^{-6x}}+e \right)}{dx}=\dfrac{d\left( {{e}^{-6x}} \right)}{d\left( -6x \right)}\dfrac{d\left( -6x \right)}{dx}+0\]
Substituting the expressions for the derivative, we get
\[\begin{align}
  & \Rightarrow \dfrac{d\left( {{e}^{-6x}}+e \right)}{dx}={{e}^{-6x}}\left( -6 \right) \\
 & \Rightarrow \dfrac{d\left( {{e}^{-6x}}+e \right)}{dx}=-6{{e}^{-6x}} \\
\end{align}\]
Thus, the derivative of the given function is \[-6{{e}^{-6x}}\].

Note: The given function has terms of the form \[{{e}^{ax}}\]. There is a special method to find the derivatives of these types of functions. We can find their derivative using the following method as: \[\dfrac{d\left( {{e}^{ax}} \right)}{dx}={{e}^{ax}}a\] .