Question

What is the second derivative of y = lnx?

Answer 1

Hint: In this question, we are given a function of x and y and we need to find its second derivative with respect to x. For this we will first find the first derivative of given function using standard derivative formula for a logarithmic function according to which ${\displaystyle \frac{d}{dx}}\ln x={\displaystyle \frac{1}{x}}$. After that, we will find derivative of the last derivative using the standard derivative formula for $x^n$ which is given as ${\displaystyle \frac{d}{dx}}{x}^n=n{x}^{n-1}$. This will give us the second derivative of the given function.

Complete step by step solution:
Here we are given the function as y = ln x. We need to find the second derivative of this function with respect to x. For this let us find the first derivative of a function before finding the second derivative.
The function is y = ln x.
Taking derivatives with respect to x on both sides of the equation we get ${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{d\ln x}{dx}}$.
According to the standard derivative formula of logarithmic function, we know that the derivative of lnx is equal to ${\displaystyle \frac{1}{x}}$. So we have ${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{1}{x}}\cdots \cdots \cdots \left(1\right)$.
This is the first derivative of y = ln x with respect to x. Now let us calculate the second derivative of y = ln x by taking the derivative of (1) with respect to x.
Taking derivative with respect to x on both sides of the equation (1) we get ${\displaystyle \frac{dy}{dx}}\left({\displaystyle \frac{d}{dx}}y\right)={\displaystyle \frac{d\left({\displaystyle \frac{1}{x}}\right)}{dx}}$.
The left side of the equation can be written as ${\displaystyle \frac{d^{2}y}{d{x}^2}}$ which denotes second derivative of y with respect to x. Solving for this right part, we know from the laws of exponent that ${\displaystyle \frac{1}{a}}$ can be written as $a^{-1}$. So we have ${\displaystyle \frac{d^{2}y}{d{x}^2}}={\displaystyle \frac{d{x}^{-1}}{dx}}$.
We know that the standard formula for the derivative of $x^n$ is given as $n{x}^{n-1}$. So let us use it to get the derivative of $x^{-1}$ we get ${\displaystyle \frac{d^{2}y}{d{x}^2}}=(-1)x^{-1-1}$.
Solving the power of x we get ${\displaystyle \frac{d^{2}y}{d{x}^2}}=-x^{-2}$.
We know that, $a^{-m}$ is equal to ${\displaystyle \frac{1}{a^m}}$. So writing the right side in same way we get $${\displaystyle \frac{d^{2}y}{d{x}^2}}={\displaystyle \frac{-1}{x^2}}$$.
This is the required second derivative of y = ln x with respect to x. Hence the final answer is $${\displaystyle \frac{d^{2}y}{d{x}^2}}={\displaystyle \frac{-1}{x^2}}$$.

Note: Students should take care of the signs while using the derivatives of $x^n$. Students should keep in mind the derivative of all basic functions. Make sure to convert the $x^{-2}$ into ${\displaystyle \frac{1}{x^2}}$ so as to get a simplified answer. Try to give the final answer as simplified as possible.