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Question: How do you differentiate \(f\left( x \right) = x\ln x - x\)?...

How do you differentiate f(x)=xln⁡x−xf\left( x \right) = x\ln x - x?

Explanation

Solution

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 ‘yy’ per unit change in ‘xx’ i.e., dydx\dfrac{{dy}}{{dx}}. The derivative can be expressed asddx\dfrac{d}{{dx}} in general. A prime dash (′)\left( ' \right)symbol is used to denote derivative of a function such as let function bef(x)f\left( x \right) then, it’s derivative is shown by f′(x)f'\left( x \right). Here, in this question we need to differentiate xln⁡x−xx\ln x - x with respect to xx.

Complete step-by-step solution:
Given the equation is, f(x)=xln⁡x−xf\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 xln⁡xx\ln xand second being xx. 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(x)f\left( x \right)is the sum of two functions i.e., u(x)u\left( x \right) and v(x)v\left( x \right)then, its derivate will be calculated as:
⇒f(x)=u(x)×v(x) ⇒f′(x)=u′(x)×v(x)+u(x)×v′(x)  \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,
⇒f(x)=xln⁡x−x ⇒f′(x)=dxdxlog⁡(x)+xd(log⁡(x))dx−ddx(x) ⇒f′(x)=log⁡(x)+x×1x−1 ⇒f′(x)=log⁡(x)+1−1 ⇒f′(x)=log⁡(x)  \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(x)=xln⁡x−xf\left( x \right) = x\ln x - xcan be differentiated intof′(x)=log⁡(x)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.dydx\dfrac{{dy}}{{dx}}is called as Leibniz’s notation.