Find `(d^(2)y)/(dx^(2))" when "y=logx+x`.

Answer

To find the second derivative \(\frac{d^2y}{dx^2}\) when \(y = \log x + x\), we will follow these steps:

### Step 1: Find the first derivative \(\frac{dy}{dx}\)

Given:
\[ y = \log x + x \]

To find \(\frac{dy}{dx}\), we differentiate \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = \frac{d}{dx}(\log x) + \frac{d}{dx}(x) \]

Using the derivatives:
- The derivative of \(\log x\) is \(\frac{1}{x}\).
- The derivative of \(x\) is \(1\).

Thus, we have:
\[ \frac{dy}{dx} = \frac{1}{x} + 1 \]

### Step 2: Simplify the first derivative

We can express \(\frac{dy}{dx}\) in a simpler form:
\[ \frac{dy}{dx} = \frac{1}{x} + 1 = \frac{1 + x}{x} \]

### Step 3: Find the second derivative \(\frac{d^2y}{dx^2}\)

Now we differentiate \(\frac{dy}{dx}\) to find \(\frac{d^2y}{dx^2}\):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{1 + x}{x}\right) \]

Using the quotient rule, where if \(u = 1 + x\) and \(v = x\), then:
\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

Calculating \(du\) and \(dv\):
- \(u = 1 + x \Rightarrow \frac{du}{dx} = 1\)
- \(v = x \Rightarrow \frac{dv}{dx} = 1\)

Applying the quotient rule:
\[ \frac{d^2y}{dx^2} = \frac{x(1) - (1 + x)(1)}{x^2} \]

### Step 4: Simplify the second derivative

Now simplify the expression:
\[ \frac{d^2y}{dx^2} = \frac{x - (1 + x)}{x^2} = \frac{x - 1 - x}{x^2} = \frac{-1}{x^2} \]

### Final Answer

Thus, the second derivative is:
\[ \frac{d^2y}{dx^2} = -\frac{1}{x^2} \]

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