
How do you differentiate $f\left( x \right) = x\ln x - x$?
Answer
542.1k+ views
Hint: This question deals with differentiation. Differentiation is a derivative of an independent variable and which can be used to calculate features in an independent variable per unit modification. Differentiation can also be said as the rate of change of ‘$y$’ per unit change in ‘$x$’ i.e., $\dfrac{{dy}}{{dx}}$. The derivative can be expressed as$\dfrac{d}{{dx}}$ in general. A prime dash $\left( ' \right)$symbol is used to denote derivative of a function such as let function be$f\left( x \right)$ then, it’s derivative is shown by $f'\left( x \right)$. Here, in this question we need to differentiate $x\ln x - x$ with respect to $x$.
Complete step-by-step solution:
Given the equation is, $f\left( x \right) = x\ln x - x$.
In order to solve this question, we first have to divide the given equation into two parts, first being $x\ln x$and second being $x$. We have to use the product rule of differentiation to solve the first part and basic differentiation to solve the second part.
The product rule says that when $f\left( x \right)$is the sum of two functions i.e., $u\left( x \right)$ and $v\left( x \right)$then, its derivate will be calculated as:
$
\Rightarrow f\left( x \right) = u\left( x \right) \times v\left( x \right) \\
\Rightarrow f'\left( x \right) = u'\left( x \right) \times v\left( x \right) + u\left( x \right) \times v'\left( x \right) \\
$
Now, using the product rule and basic differentiation in the given equation, we get,
$
\Rightarrow f\left( x \right) = x\ln x - x \\
\Rightarrow f'\left( x \right) = \dfrac{{dx}}{{dx}}\log \left( x \right) + x\dfrac{{d\left( {\log \left( x \right)} \right)}}{{dx}} - \dfrac{d}{{dx}}\left( x \right) \\
\Rightarrow f'\left( x \right) = \log \left( x \right) + x \times \dfrac{1}{x} - 1 \\
\Rightarrow f'\left( x \right) = \log \left( x \right) + 1 - 1 \\
\Rightarrow f'\left( x \right) = \log \left( x \right) \\
$
Hence, $f\left( x \right) = x\ln x - x$can be differentiated into$f'\left( x \right) = \log \left( x \right)$.
Note: This problem and many others similar to this can be easily solved using different rules of differentiation. Some of the most commonly used differentiation rules are sum or difference rule, product rule, quotient rule and chain rule. The list for differentiation formulas includes derivatives for constants, polynomials, trigonometric functions and many more.$\dfrac{{dy}}{{dx}}$is called as Leibniz’s notation.
Complete step-by-step solution:
Given the equation is, $f\left( x \right) = x\ln x - x$.
In order to solve this question, we first have to divide the given equation into two parts, first being $x\ln x$and second being $x$. We have to use the product rule of differentiation to solve the first part and basic differentiation to solve the second part.
The product rule says that when $f\left( x \right)$is the sum of two functions i.e., $u\left( x \right)$ and $v\left( x \right)$then, its derivate will be calculated as:
$
\Rightarrow f\left( x \right) = u\left( x \right) \times v\left( x \right) \\
\Rightarrow f'\left( x \right) = u'\left( x \right) \times v\left( x \right) + u\left( x \right) \times v'\left( x \right) \\
$
Now, using the product rule and basic differentiation in the given equation, we get,
$
\Rightarrow f\left( x \right) = x\ln x - x \\
\Rightarrow f'\left( x \right) = \dfrac{{dx}}{{dx}}\log \left( x \right) + x\dfrac{{d\left( {\log \left( x \right)} \right)}}{{dx}} - \dfrac{d}{{dx}}\left( x \right) \\
\Rightarrow f'\left( x \right) = \log \left( x \right) + x \times \dfrac{1}{x} - 1 \\
\Rightarrow f'\left( x \right) = \log \left( x \right) + 1 - 1 \\
\Rightarrow f'\left( x \right) = \log \left( x \right) \\
$
Hence, $f\left( x \right) = x\ln x - x$can be differentiated into$f'\left( x \right) = \log \left( x \right)$.
Note: This problem and many others similar to this can be easily solved using different rules of differentiation. Some of the most commonly used differentiation rules are sum or difference rule, product rule, quotient rule and chain rule. The list for differentiation formulas includes derivatives for constants, polynomials, trigonometric functions and many more.$\dfrac{{dy}}{{dx}}$is called as Leibniz’s notation.
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