Question

If \( y + x e^y = \sin x + \tan x \), then the value of \( \frac{dy}{dx} \) at \( x = 0 \) is ...................

Hint:

When differentiating implicitly, remember to use the product rule and chain rule when necessary. For evaluating at specific values of \( x \), substitute the values directly into the equation.

Answer 1

We are given the equation: \[ y + x e^y = \sin x + \tan x \] We need to find \( \frac{dy}{dx} \) at \( x = 0 \).
Step 1: Differentiate both sides implicitly with respect to \( x \).
On the left-hand side:
- \( \frac{d}{dx}(y) = \frac{dy}{dx} \)
- Apply the product rule to \( x e^y \): \[ \frac{d}{dx}(x e^y) = e^y + x e^y \frac{dy}{dx} \] Thus, differentiating the left-hand side: \[ \frac{d}{dx}\left( y + x e^y \right) = \frac{dy}{dx} + e^y + x e^y \frac{dy}{dx} \] On the right-hand side:
- \( \frac{d}{dx}(\sin x) = \cos x \)
- \( \frac{d}{dx}(\tan x) = \sec^2 x \)
Thus, differentiating the right-hand side: \[ \frac{d}{dx}\left( \sin x + \tan x \right) = \cos x + \sec^2 x \] Step 2: Substitute the derivatives.
Now we equate both sides: \[ \frac{dy}{dx} + e^y + x e^y \frac{dy}{dx} = \cos x + \sec^2 x \] Step 3: Evaluate at \( x = 0 \).
At \( x = 0 \), we know:
- \( \sin(0) = 0 \), \( \tan(0) = 0 \)
- \( \cos(0) = 1 \), \( \sec^2(0) = 1 \)
- Also, from the original equation \( y + x e^y = \sin x + \tan x \), at \( x = 0 \), we get: \[ y + 0 \cdot e^y = 0 \quad \Rightarrow \quad y = 0 \] Substitute \( y = 0 \) and \( x = 0 \) into the derivative equation: \[ \frac{dy}{dx} + e^0 + 0 \cdot e^0 \cdot \frac{dy}{dx} = 1 + 1 \] \[ \frac{dy}{dx} + 1 = 2 \] Thus, solving for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = 2 - 1 = 1 \] Final Answer: \[ \boxed{1} \]