If `s=sqrt(t^(2)+1)`, then `(d^(2)s)/(dt^(2))` is equal to

Options

A. A) `1/s`
B. B) `(1)/(s^(2))`
C. C) `(1)/(s^(3))`
D. D) `(1)/(s^(4))`

Answer

To find the second derivative of \( s \) with respect to \( t \) given that \( s = \sqrt{t^2 + 1} \), we will follow these steps:

### Step 1: Express \( s \) in terms of \( t \)
We start with the equation:
\[ s = \sqrt{t^2 + 1} \]

### Step 2: Square both sides
Squaring both sides gives us:
\[ s^2 = t^2 + 1 \]
This can be rearranged to:
\[ s^2 - t^2 = 1 \tag{1} \]

### Step 3: Differentiate both sides with respect to \( t \)
Differentiating equation (1) with respect to \( t \) using implicit differentiation:
\[ \frac{d}{dt}(s^2) - \frac{d}{dt}(t^2) = \frac{d}{dt}(1) \]
This results in:
\[ 2s \frac{ds}{dt} - 2t = 0 \]
Simplifying this gives:
\[ 2s \frac{ds}{dt} = 2t \]
Dividing both sides by 2:
\[ s \frac{ds}{dt} = t \tag{2} \]

### Step 4: Differentiate equation (2) to find the second derivative
Now we differentiate equation (2) with respect to \( t \):
\[ \frac{d}{dt}(s \frac{ds}{dt}) = \frac{d}{dt}(t) \]
Using the product rule on the left side:
\[ \frac{ds}{dt} \cdot \frac{ds}{dt} + s \cdot \frac{d^2s}{dt^2} = 1 \]
This can be rewritten as:
\[ \left(\frac{ds}{dt}\right)^2 + s \frac{d^2s}{dt^2} = 1 \]

### Step 5: Substitute \( \frac{ds}{dt} \) from equation (2)
From equation (2), we have \( \frac{ds}{dt} = \frac{t}{s} \). Substituting this into our equation gives:
\[ \left(\frac{t}{s}\right)^2 + s \frac{d^2s}{dt^2} = 1 \]
This simplifies to:
\[ \frac{t^2}{s^2} + s \frac{d^2s}{dt^2} = 1 \]

### Step 6: Solve for \( \frac{d^2s}{dt^2} \)
Rearranging the equation:
\[ s \frac{d^2s}{dt^2} = 1 - \frac{t^2}{s^2} \]
Dividing both sides by \( s \):
\[ \frac{d^2s}{dt^2} = \frac{1}{s} - \frac{t^2}{s^3} \]
Combining the terms gives:
\[ \frac{d^2s}{dt^2} = \frac{s^2 - t^2}{s^3} \]

### Step 7: Substitute \( s^2 - t^2 \) using equation (1)
From equation (1), we know that \( s^2 - t^2 = 1 \):
\[ \frac{d^2s}{dt^2} = \frac{1}{s^3} \]

### Final Result
Thus, the second derivative of \( s \) with respect to \( t \) is:
\[ \frac{d^2s}{dt^2} = \frac{1}{s^3} \]

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