Question

How do you find the derivative of \[y = x.\ln x\] ?

Answer 1

Hint: Derivatives are defined as the varying rate of a function with respect to an independent variable. Since they ask us to find only differentiation of ‘y’ with respect to ‘x’. To solve this we use the product rule. That is if we have \[y = uv\] then its differentiation with respect to ‘x’ is \[\dfrac{{dy}}{{dx}} = u \times \dfrac{{dv}}{{dx}} + v \times \dfrac{{du}}{{dx}}\] .

Complete step-by-step answer:
Given, \[y = x.\ln x\]
We know the product rule of a differentiation is
\[\dfrac{{dy}}{{dx}} = u \times \dfrac{{dv}}{{dx}} + v \times \dfrac{{du}}{{dx}}\] .
Here, \[u = x\] and \[v = \ln x\] .
As we can see need
\[\dfrac{{dv}}{{dx}} = \dfrac{{d(\ln x)}}{{dx}} = \dfrac{1}{x}\] .
Because we know that differentiation of \[\ln x\] with respect to ‘x’ is \[\dfrac{1}{x}\] .
We need
 \[\dfrac{{du}}{{dx}} = \dfrac{{d(x)}}{{dx}} = 1\]
Now \[y = x.\ln x\] , differentiating with respect to ‘x’ we have,
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = x \times \dfrac{{d(\ln x)}}{{dx}} + \ln x \times \dfrac{{d(x)}}{{dx}}\]
Substituting the values we have,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = x \times \dfrac{1}{x} + \ln x \times 1\]
 Cancelling and simplifying we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = 1 + \ln x\] Is the required result.
So, the correct answer is “ \[ \dfrac{{dy}}{{dx}} = 1 + \ln x\] ”.

Note: We know the differentiation of \[{x^n}\] is \[\dfrac{{d({x^n})}}{{dx}} = n.{x^{n - 1}}\] . The obtained result is the first derivative. If we differentiate again we get a second derivative. If we differentiate the second derivative again we get a third derivative and so on. Careful in applying the product rule. We also know that differentiation of constant terms is zero.