If ` x = t + (1)/(t) "and " y = t - (1)/(t) . "then" (dy)/(dx) ` is equal to

Options

A. `(2g)/((t^(2)+1))`
B. `((t^(2)+1))/((t^(2)-1))`
C. `((t^(2)-1))/((t^(2)-1))`
D. `(2g)/((1-t^(2)))`

Answer

To find \(\frac{dy}{dx}\) given the equations \(x = t + \frac{1}{t}\) and \(y = t - \frac{1}{t}\), we will follow these steps:

### Step 1: Differentiate \(x\) with respect to \(t\)

Given:
\[ x = t + \frac{1}{t} \]

Differentiating \(x\) with respect to \(t\):
\[ \frac{dx}{dt} = \frac{d}{dt}(t) + \frac{d}{dt}\left(\frac{1}{t}\right) \]
Using the power rule and the derivative of \(\frac{1}{t}\):
\[ \frac{dx}{dt} = 1 - \frac{1}{t^2} \]

### Step 2: Differentiate \(y\) with respect to \(t\)

Given:
\[ y = t - \frac{1}{t} \]

Differentiating \(y\) with respect to \(t\):
\[ \frac{dy}{dt} = \frac{d}{dt}(t) - \frac{d}{dt}\left(\frac{1}{t}\right) \]
Using the power rule and the derivative of \(\frac{1}{t}\):
\[ \frac{dy}{dt} = 1 + \frac{1}{t^2} \]

### Step 3: Find \(\frac{dy}{dx}\)

Using the chain rule, we can express \(\frac{dy}{dx}\) as:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]

Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{1 + \frac{1}{t^2}}{1 - \frac{1}{t^2}} \]

### Step 4: Simplify the expression

To simplify:
\[ \frac{dy}{dx} = \frac{1 + \frac{1}{t^2}}{1 - \frac{1}{t^2}} = \frac{\frac{t^2 + 1}{t^2}}{\frac{t^2 - 1}{t^2}} = \frac{t^2 + 1}{t^2 - 1} \]

### Final Answer

Thus, we have:
\[ \frac{dy}{dx} = \frac{t^2 + 1}{t^2 - 1} \]

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