Math Problem Statement

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

Solution

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

First, we define the intermediate variables:

  • x(t)=etx(t) = e^t
  • y(t)=cos(t)y(t) = \cos(t)

The function ff in terms of tt is: f(t)=sin(et+2cos(t))f(t) = \sin(e^t + 2\cos(t))

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

Using the chain rule: dfdt=fxdxdt+fydydt\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 ff: f(x,y)=sin(x+2y)f(x, y) = \sin(x + 2y)

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

Next, compute the derivatives of x(t)x(t) and y(t)y(t) with respect to tt: dxdt=ddtet=et\frac{dx}{dt} = \frac{d}{dt} e^t = e^t dydt=ddtcos(t)=sin(t)\frac{dy}{dt} = \frac{d}{dt} \cos(t) = -\sin(t)

Now, substitute these into the chain rule expression: dfdt=cos(x+2y)dxdt+2cos(x+2y)dydt\frac{df}{dt} = \cos(x + 2y) \cdot \frac{dx}{dt} + 2\cos(x + 2y) \cdot \frac{dy}{dt} dfdt=cos(et+2cos(t))et+2cos(et+2cos(t))(sin(t))\frac{df}{dt} = \cos(e^t + 2\cos(t)) \cdot e^t + 2\cos(e^t + 2\cos(t)) \cdot (-\sin(t)) dfdt=cos(et+2cos(t))et2cos(et+2cos(t))sin(t)\frac{df}{dt} = \cos(e^t + 2\cos(t)) \cdot e^t - 2\cos(e^t + 2\cos(t)) \cdot \sin(t) dfdt=cos(et+2cos(t))(et2sin(t))\frac{df}{dt} = \cos(e^t + 2\cos(t)) (e^t - 2\sin(t))

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

However, given the expression derived, the correct selection should be: b.cos(et+2cos(t))(et2sint)\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.

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Math Problem Analysis

Mathematical Concepts

Chain rule
Partial derivatives
Trigonometric functions

Formulas

Chain rule formula for multivariable functions

Theorems

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Suitable Grade Level

Advanced Calculus