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Question

Question: What is the second derivative of y = lnx?...

What is the second derivative of y = lnx?

Explanation

Solution

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 ddxlnx=1x\dfrac{d}{dx}\ln x=\dfrac{1}{x}. After that, we will find derivative of the last derivative using the standard derivative formula for xn{{x}^{n}} which is given as ddxxn=nxn1\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 dydx=dlnxdx\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 1x\dfrac{1}{x}. So we have dydx=1x(1)\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 dydx(ddxy)=d(1x)dx\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 d2ydx2\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 1a\dfrac{1}{a} can be written as a1{{a}^{-1}}. So we have d2ydx2=dx1dx\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d{{x}^{-1}}}{dx}.
We know that the standard formula for the derivative of xn{{x}^{n}} is given as nxn1n{{x}^{n-1}}. So let us use it to get the derivative of x1{{x}^{-1}} we get d2ydx2=(1)x11\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\left( -1 \right){{x}^{-1-1}}.
Solving the power of x we get d2ydx2=x2\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-{{x}^{-2}}.
We know that, am{{a}^{-m}} is equal to 1am\dfrac{1}{{{a}^{m}}}. So writing the right side in same way we get d2ydx2=1x2\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 d2ydx2=1x2\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-1}{{{x}^{2}}}.

Note: Students should take care of the signs while using the derivatives of xn{{x}^{n}}. Students should keep in mind the derivative of all basic functions. Make sure to convert the x2{{x}^{-2}} into 1x2\dfrac{1}{{{x}^{2}}} so as to get a simplified answer. Try to give the final answer as simplified as possible.