Find the second derivatives of the following functions :
`logx`

Answer

To find the second derivative of the function \( y = \log x \), we will follow these steps:

### Step 1: Find the first derivative
We start with the function:
\[ y = \log x \]
To find the first derivative, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(\log x) \]
Using the derivative rule for logarithmic functions, we know that:
\[ \frac{d}{dx}(\log x) = \frac{1}{x} \]
Thus, the first derivative is:
\[ \frac{dy}{dx} = \frac{1}{x} \]

### Step 2: Find the second derivative
Next, we need to find the second derivative. We differentiate the first derivative \( \frac{dy}{dx} \) with respect to \( x \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{1}{x}\right) \]
We can rewrite \( \frac{1}{x} \) as \( x^{-1} \). Now we differentiate:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(x^{-1}) \]
Using the power rule for differentiation, where \( \frac{d}{dx}(x^m) = mx^{m-1} \):
\[ \frac{d}{dx}(x^{-1}) = -1 \cdot x^{-1-1} = -1 \cdot x^{-2} = -\frac{1}{x^2} \]
Thus, the second derivative is:
\[ \frac{d^2y}{dx^2} = -\frac{1}{x^2} \]

### Final Answer
The second derivative of \( \log x \) is:
\[ \frac{d^2y}{dx^2} = -\frac{1}{x^2} \]

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