Question

Find the second order derivative of x (log x).

Answer 1

Hint: Derivative means to differentiate the given term. Here the second derivative has been asked which means we have to differentiate two times. Also we will be using the product rule of differentiation.

Complete step-by-step answer:
Derivatives are defined as the varying rate of change of a function with respect to an independent variable. The derivative is primarily used when there is some varying quantity, and the rate of change is not constant. The derivative is used to measure the sensitivity of one variable (dependent variable) with respect to another variable (independent variable).
The second-order derivatives are also used to get an idea of the shape of the graph for the given function.
Product rule: If $y=uv$ then its differentiation is $\dfrac{dy}{dx}=u\times \dfrac{dv}{dx}+v\times \dfrac{du}{dx}$.
Let $y=x\,\log x.......(1)$
Applying product rule on equation (1) as $u$ is $x$ and $v$ is $\log x$ and getting the first derivative with respect to $x$ and also differentiation of $x$ is $1$ and differentiation of $\log x$ is $\dfrac{1}{x}$. So we get,
$\,\dfrac{dy}{dx}=x\times \dfrac{1}{x}+\log x\times 1........(2)$
Now cancelling similar terms in equation (2) we get,
$\,\dfrac{dy}{dx}=1+\log x......(3)$
Again differentiating equation (3) to get the second order derivative with respect to $x$
$\dfrac{d}{dx}\,\left( \dfrac{dy}{dx} \right)=\dfrac{d\left( 1+\log x \right)}{dx}.........(4)$
Rearranging equation (4) and differentiating the terms on right hand side we get,

& \,\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d(1)}{dx}+\dfrac{d(\log x)}{dx} \\\ & \,\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=0+\dfrac{1}{x}=\dfrac{1}{x} \\\ \end{aligned}$$ (Differentiation of a constant is always zero i.e. $\dfrac{d(1)}{dx}=0$) Hence the second order derivative of $$x\,\log x$$ is $$\dfrac{1}{x}$$. Note: We need to understand that derivative means differentiation. Also we need to remember the differentiation of x and log x. We can commit a mistake in applying product rule by interchanging u and v in a hurry. Also differentiation of a constant is always zero.