Question

What is the derivative of ${x^e}$ ?

Answer 1

Hint: In this question, we need to differentiate the given function ${x^e}$ with respect to the variable $x$. Differentiation can be defined as a derivative of independent variable value and can be used to calculate features in an independent variable.Note that for the given function, we can use the power rule of differentiation $\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$. So find the value of $n$ and simplify using the formula to obtain the derivative of the given function.

Complete step by step answer:
We have to evaluate the derivative in variable x ${x^e}$ using the power rule of differentiation. So, to evaluate the derivative, we have to differentiate the function with respect to $x$. We will be using the power rule of differentiation to evaluate the derivative of the given function. So, we have, $f(x) = {x^e}$. Differentiating both sides with respect to the variable $t$, we get,
$ \Rightarrow f'(x) = \dfrac{d}{{dx}}\left( {{x^e}} \right)$

Now, we notice that our function resembles the term ${x^n}$. So, we can apply the power rule of differentiation directly to find the derivative of the given function. So, by comparing ${x^n}$ with ${x^e}$, we get the value of $n$ as $e$.Now, using the power rule of differentiation,
$\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$, we get,
$ \Rightarrow f'(x) = e\left( {{x^{e - 1}}} \right)$
Simplifying the expression, we get,
$ \therefore f'(x) = e \times {x^{e - 1}}$


Note: The derivative of a constant is always equal to zero and we can take out the coefficients of the terms outside of the differentiation as $\dfrac{{d\left( {k \times f\left( x \right)} \right)}}{{dx}} = k \times \dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}}$. If the function is a sum of two or more terms, we can differentiate both the terms separately as $\dfrac{{d\left( {f\left( x \right) + g\left( x \right)} \right)}}{{dx}} = \dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}} + \dfrac{{d\left( {g\left( x \right)} \right)}}{{dx}}$.