Question

Let \( f(x, y) = e^x \sin y, \, x = t^3 + 1, \, y = t^4 + t. \) Then \( \dfrac{df}{dt} \) at \( t = 0 \) is ............. (rounded off to two decimal places).

Hint:

For composite functions \( f(x(t), y(t)) \), use the chain rule with both partials. Always evaluate at the given \( t \).

Answer 1

Correct Answer: 2.7

Step 1: Chain rule for partial derivatives.
\[ \frac{df}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt}. \]

Step 2: Compute partial derivatives.
\[ \frac{\partial f}{\partial x} = e^x \sin y, \frac{\partial f}{\partial y} = e^x \cos y. \] Also, \[ \frac{dx}{dt} = 3t^2, \frac{dy}{dt} = 4t^3 + 1. \]

Step 3: Substitute and evaluate at \( t = 0. \)
At \( t = 0 \), \( x = 1, y = 0. \) \[ \frac{df}{dt} = e^1 \sin 0 (3(0)^2) + e^1 \cos 0 (1) = e \times 1 = e. \] Rounded to two decimal places, \( e \approx 2.72. \)

Final Answer: \[ \boxed{2.72} \]