If ` t (1 +x^2) =x and x^2 +t^2 = y ` then at ` x=2 ` the value of `(dy)/(dx)` is

Options

A. `488/125`
B. `88/125`
C. `101/125`
D. None

Answer

To solve the problem step by step, we start with the given equations:

1. **Equations Given:**
\[ t(1 + x^2) = x \quad \text{(1)} \]
\[ y = x^2 + t^2 \quad \text{(2)} \]

2. **Differentiate Equation (2) with respect to \(x\):**
We need to find \(\frac{dy}{dx}\). Using the chain rule, we differentiate \(y\):
\[ \frac{dy}{dx} = \frac{d}{dx}(x^2) + \frac{d}{dx}(t^2) \]
\[ \frac{dy}{dx} = 2x + 2t \frac{dt}{dx} \quad \text{(3)} \]

3. **Find \(t\) from Equation (1):**
Rearranging Equation (1) to solve for \(t\):
\[ t = \frac{x}{1 + x^2} \]

4. **Differentiate \(t\) with respect to \(x\):**
We need to find \(\frac{dt}{dx}\):
Using the quotient rule:
\[ \frac{dt}{dx} = \frac{(1 + x^2)(1) - x(2x)}{(1 + x^2)^2} \]
Simplifying:
\[ \frac{dt}{dx} = \frac{1 + x^2 - 2x^2}{(1 + x^2)^2} = \frac{1 - x^2}{(1 + x^2)^2} \quad \text{(4)} \]

5. **Substitute \(t\) and \(\frac{dt}{dx}\) back into Equation (3):**
Now we substitute \(t\) and \(\frac{dt}{dx}\) into the expression for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = 2x + 2\left(\frac{x}{1 + x^2}\right)\left(\frac{1 - x^2}{(1 + x^2)^2}\right) \]
\[ = 2x + \frac{2x(1 - x^2)}{(1 + x^2)^3} \quad \text{(5)} \]

6. **Evaluate \(\frac{dy}{dx}\) at \(x = 2\):**
Substitute \(x = 2\) into Equation (5):
\[ \frac{dy}{dx} = 2(2) + \frac{2(2)(1 - 2^2)}{(1 + 2^2)^3} \]
\[ = 4 + \frac{2(2)(1 - 4)}{(1 + 4)^3} \]
\[ = 4 + \frac{4(-3)}{5^3} \]
\[ = 4 - \frac{12}{125} \]
\[ = \frac{500}{125} - \frac{12}{125} = \frac{488}{125} \]

7. **Final Answer:**
The value of \(\frac{dy}{dx}\) at \(x = 2\) is:
\[ \frac{dy}{dx} = \frac{488}{125} \]