
How do you find the derivative of \[{{e}^{-6x}}+e\]?
Answer
532.8k+ views
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: \[\dfrac{d\left( {{e}^{ax}} \right)}{dx}={{e}^{ax}}a\] .
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: \[\dfrac{d\left( {{e}^{ax}} \right)}{dx}={{e}^{ax}}a\] .
Recently Updated Pages
A man running at a speed 5 ms is viewed in the side class 12 physics CBSE

The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

State and explain Hardy Weinbergs Principle class 12 biology CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Which of the following statements is wrong a Amnion class 12 biology CBSE

Differentiate between action potential and resting class 12 biology CBSE

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

Explain sex determination in humans with the help of class 12 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

