Q.Find the second order derivative of the function: $\log x$

The key idea is that the second derivative of $\log x$ is found by differentiating the first derivative, using the standard derivative of the natural logarithm.

Step 1: Let $y = \log x$ (natural logarithm). The first derivative is:

Step 2: Differentiate $\frac{dy}{dx}$ with respect to $x$ to get the second derivative:

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left( \frac{1}{x} \right)$ …

We are asked for the second order derivative of $\log x$. In calculus, when the base of the logarithm is not specified, and the context is standard differentiation, $\log x$ almost always means the natural logarithm, $\ln x$. This is the convention in most Indian exam syllabi (CBSE, JEE, etc.). If the base were 10, it would usually be written as $\log_{10} x$ or explicitly stated.

The core idea is straightforward: the second derivative is simply the derivative of the first derivative. So we differentiate once, then differentiate that result.

1. First derivative of $\log x$

The derivative of $\ln x$ with respect to $x$ is a fundamental result:

This comes from the definition of the natural logarithm as the inverse of the exponential function. If $y = \ln x$, then $e^y = x$, and differentiating implicitly gives $e^y \frac{dy}{dx} = 1$, so $\frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}$.

2. Second derivative

Now we differentiate $\frac{1}{x}$ with respect to $x$. It is often easier to rewrite $\frac{1}{x}$ as $x^{-1}$ before differentiating.

Using the power rule $\frac{d}{dx}(x^n) = n x^{n-1}$:

$\frac{d}{dx} (x^{-1}) = (-1) x^{-1-1} = -x^{-2}$ …

Method: Successive Differentiation of a Logarithmic Function

This method finds a higher-order derivative of a natural-logarithm expression by first reducing it to its known first-derivative form, then differentiating that result using the power rule.

Steps

Step 1: Recall the standard derivative of the logarithm

For the natural logarithm, memorise the base result:

This single fact converts a logarithmic differentiation problem into an ordinary power-rule problem after the first step.

Step 2: Rewrite the first derivative as a power of $x$

Express $\dfrac{1}{x}$ as $x^{-1}$ so that the power rule can be applied directly for the next differentiation. …

Common Mistakes

Mistake 1: Dropping the negative sign when differentiating $\dfrac{1}{x}$

Why it's wrong: rewriting $\frac{1}{x}$ as $x^{-1}$ and applying the power rule gives $(-1)x^{-2} = -\frac{1}{x^2}$ — the negative sign comes directly from the negative exponent and is easy to drop by accident. Correct approach: always carry the exponent's sign through explicitly, and sanity-check that $1/x$ has negative slope for $x>0$, so its derivative must be negative.

Mistake 2: Ignoring the domain restriction $x>0$

Why it's wrong: $\log x$ (natural log) is only defined for positive $x$, so both its first and second derivatives inherit that same restriction — stating the answer as valid "for all $x$" is technically incorrect. Correct approach: report $\dfrac{d^2y}{dx^2}=-\dfrac{1}{x^2}$ as valid for $x>0$ only. …