
How do you differentiate \[x{{e}^{x}}\]?
Answer
541.5k+ views
Hint: Assume the given function as \[f\left( x \right)\]. Consider \[f\left( x \right)\] as the product of an algebraic function and an exponential function. Now, apply the product rule of differentiation given as: - \[\dfrac{d\left( u\times v \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\]. Here, consider, u = x and \[v={{e}^{x}}\]. Use the formula: - \[\dfrac{d{{e}^{x}}}{dx}={{e}^{x}}\] to simplify the derivative and get the answer.
Complete step by step answer:
Here, we have been provided with the function \[x{{e}^{x}}\] and we are asked to differentiate it. Let us assume the given function as \[f\left( x \right)\]. So, we have,
\[\Rightarrow f\left( x \right)=x{{e}^{x}}\]
Now, we can assume the given function as the product of an algebraic function (x) and an exponential function (\[{{e}^{x}}\]). So, we have,
\[\Rightarrow f\left( x \right)=x\times {{e}^{x}}\]
Let us assume x and \[{{e}^{x}}\] as ‘u’ and ‘v’ respectively. So, we have,
\[\Rightarrow f\left( x \right)=u\times v\]
Differentiating both sides with respect to x, we get,
\[\Rightarrow \dfrac{df\left( x \right)}{dx}=\dfrac{d\left( u\times v \right)}{dx}\]
Now, applying the product rule of differentiation given as: -
\[\Rightarrow \dfrac{d\left( u\times v \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\], we get,
\[\Rightarrow \dfrac{df\left( x \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\]
Substituting the assumed values of u and v, we get,
\[\Rightarrow \dfrac{df\left( x \right)}{dx}=x\dfrac{d{{e}^{x}}}{dx}+{{e}^{x}}\dfrac{dx}{dx}\]
We know that \[\dfrac{d{{e}^{x}}}{dx}={{e}^{x}}\], so we have,
\[\begin{align}
& \Rightarrow \dfrac{df\left( x \right)}{dx}=x{{e}^{x}}+{{e}^{x}}\times 1 \\
& \Rightarrow \dfrac{df\left( x \right)}{dx}={{e}^{x}}\left( x+1 \right) \\
\end{align}\]
Hence, the above relation is our answer.
Note:
One may note that whenever we are asked to differentiate a product of two or more functions we apply the product rule. You must remember all the basic rules and formulas of differentiation like - product rule, chain rule, \[\dfrac{u}{v}\] rule, etc. as they make our question easy to solve. Remember the derivatives of some common functions like: - \[{{x}^{n}},{{e}^{x}}\], trigonometric functions, logarithmic functions etc as they are used frequently in calculus.
Complete step by step answer:
Here, we have been provided with the function \[x{{e}^{x}}\] and we are asked to differentiate it. Let us assume the given function as \[f\left( x \right)\]. So, we have,
\[\Rightarrow f\left( x \right)=x{{e}^{x}}\]
Now, we can assume the given function as the product of an algebraic function (x) and an exponential function (\[{{e}^{x}}\]). So, we have,
\[\Rightarrow f\left( x \right)=x\times {{e}^{x}}\]
Let us assume x and \[{{e}^{x}}\] as ‘u’ and ‘v’ respectively. So, we have,
\[\Rightarrow f\left( x \right)=u\times v\]
Differentiating both sides with respect to x, we get,
\[\Rightarrow \dfrac{df\left( x \right)}{dx}=\dfrac{d\left( u\times v \right)}{dx}\]
Now, applying the product rule of differentiation given as: -
\[\Rightarrow \dfrac{d\left( u\times v \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\], we get,
\[\Rightarrow \dfrac{df\left( x \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\]
Substituting the assumed values of u and v, we get,
\[\Rightarrow \dfrac{df\left( x \right)}{dx}=x\dfrac{d{{e}^{x}}}{dx}+{{e}^{x}}\dfrac{dx}{dx}\]
We know that \[\dfrac{d{{e}^{x}}}{dx}={{e}^{x}}\], so we have,
\[\begin{align}
& \Rightarrow \dfrac{df\left( x \right)}{dx}=x{{e}^{x}}+{{e}^{x}}\times 1 \\
& \Rightarrow \dfrac{df\left( x \right)}{dx}={{e}^{x}}\left( x+1 \right) \\
\end{align}\]
Hence, the above relation is our answer.
Note:
One may note that whenever we are asked to differentiate a product of two or more functions we apply the product rule. You must remember all the basic rules and formulas of differentiation like - product rule, chain rule, \[\dfrac{u}{v}\] rule, etc. as they make our question easy to solve. Remember the derivatives of some common functions like: - \[{{x}^{n}},{{e}^{x}}\], trigonometric functions, logarithmic functions etc as they are used frequently in calculus.
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