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 $\dfrac{d}{dx}\ln x=\dfrac{1}{x}$. After that, we will find derivative of the last derivative using the standard derivative formula for ${{x}^{n}}$ which is given as $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$. 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 $\dfrac{dy}{dx}=\dfrac{d\ln x}{dx}$.
According to the standard derivative formula of logarithmic function, we know that the derivative of lnx is equal to $\dfrac{1}{x}$. So we have $\dfrac{dy}{dx}=\dfrac{1}{x}\cdots \cdots \cdots \left( 1 \right)$.
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 $\dfrac{dy}{dx}\left( \dfrac{d}{dx}y \right)=\dfrac{d\left( \dfrac{1}{x} \right)}{dx}$.
The left side of the equation can be written as $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ which denotes second derivative of y with respect to x. Solving for this right part, we know from the laws of exponent that $\dfrac{1}{a}$ can be written as ${{a}^{-1}}$. So we have $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d{{x}^{-1}}}{dx}$.
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 $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\left( -1 \right){{x}^{-1-1}}$.
Solving the power of x we get $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-{{x}^{-2}}$.
We know that, ${{a}^{-m}}$ is equal to $\dfrac{1}{{{a}^{m}}}$. So writing the right side in same way we get \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-1}{{{x}^{2}}}\].
This is the required second derivative of y = ln x with respect to x. Hence the final answer is \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-1}{{{x}^{2}}}\].

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 $\dfrac{1}{{{x}^{2}}}$ so as to get a simplified answer. Try to give the final answer as simplified as possible.