Consider the function $f(x,y)=\sin(x+2y)$ where $x(t)=e^t$ and $y(t)=\cos(t)$. The derivative of f with respect to t equals


a.None of these  

b.$\cos(e^t+2\cos(t))(e^t-2\sin t)$  

c. $\cos(e^t+2\cos(t))(e^t-2)$

d.$\cos(e^t+2\cos(t))(e^t+2\sin t)$

e.$\cos(e^t-2\cos(t))(1-2\sin t)$

To find the derivative of the function $ f(x, y) = \sin(x + 2y) $ with respect to $ t $, where $ x(t) = e^t $ and $ y(t) = \cos(t) $, we will use the chain rule for partial derivatives.

First, we define the intermediate variables:
- $ x(t) = e^t $
- $ y(t) = \cos(t) $

The function $ f $ in terms of $ t $ is:
\[ f(t) = \sin(e^t + 2\cos(t)) \]

We need to find $ \frac{df}{dt} $.

Using the chain rule:
\[ \frac{df}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} \]

First, compute the partial derivatives of $ f $:
\[ f(x, y) = \sin(x + 2y) \]

Thus,
\[ \frac{\partial f}{\partial x} = \cos(x + 2y) \]
\[ \frac{\partial f}{\partial y} = 2\cos(x + 2y) \]

Next, compute the derivatives of $ x(t) $ and $ y(t) $ with respect to $ t $:
\[ \frac{dx}{dt} = \frac{d}{dt} e^t = e^t \]
\[ \frac{dy}{dt} = \frac{d}{dt} \cos(t) = -\sin(t) \]

Now, substitute these into the chain rule expression:
\[ \frac{df}{dt} = \cos(x + 2y) \cdot \frac{dx}{dt} + 2\cos(x + 2y) \cdot \frac{dy}{dt} \]
\[ \frac{df}{dt} = \cos(e^t + 2\cos(t)) \cdot e^t + 2\cos(e^t + 2\cos(t)) \cdot (-\sin(t)) \]
\[ \frac{df}{dt} = \cos(e^t + 2\cos(t)) \cdot e^t - 2\cos(e^t + 2\cos(t)) \cdot \sin(t) \]
\[ \frac{df}{dt} = \cos(e^t + 2\cos(t)) (e^t - 2\sin(t)) \]

Therefore, the correct answer is:
\[ \boxed{d. \cos(e^t + 2\cos(t)) (e^t + 2\sin t)} \]

However, given the expression derived, the correct selection should be:
\[ \boxed{b. \cos(e^t + 2\cos(t)) (e^t - 2\sin t)} \]

Would you like further details or have any questions?

Here are 5 questions you might ask next:
1. Can you explain the chain rule in multivariable calculus in more detail?
2. How do you apply the product rule when differentiating a product of functions?
3. What is the significance of partial derivatives in this context?
4. Can you show another example of using the chain rule for different functions?
5. How do you interpret the result of the derivative in terms of the original function's behavior?

**Tip:** When applying the chain rule in multivariable calculus, always ensure to keep track of each intermediate function and its dependence on the original variables.

Understand how to find the derivative of the function sin(x + 2y) with respect to t, given x(t) = e^t and y(t) = cos(t). Includes step-by-step solution using the chain rule for partial derivatives. Suitable for advanced calculus students and anyone interested in multivariable calculus concepts.

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