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Magazine Chapter 9: Problem 52
Use (8) to find the indicated derivative. $$ w=e^{x y} ; x=\frac{4}{2 t+1}, y=3 t+5 ;\left.\frac{d w}{d t}\right|_{t=0} $$
Short Answer
Expert verified
The derivative \( \frac{d w}{d t} \) at \( t = 0 \) is \( -28 e^{20} \).
Step by step solution
01
Understand the Function Composition
The function we are working with is given as \( w = e^{x y} \), where \( x = \frac{4}{2t+1} \) and \( y = 3t + 5 \). This means that \( w \) depends on both \( x \) and \( y \), which in turn depend on \( t \). To find the derivative of \( w \) with respect to \( t \), we'll use the chain rule.
02
Apply the Chain Rule
We'll apply the chain rule to find \( \frac{dw}{dt} \) using the multi-variable derivative chain rule. The formula for the chain rule in this context is: \[ \frac{d w}{d t} = \frac{\partial w}{\partial x} \cdot \frac{d x}{d t} + \frac{\partial w}{\partial y} \cdot \frac{d y}{d t} \].
03
Compute Partial Derivatives
First, calculate the partial derivatives: - \( \frac{\partial w}{\partial x} = \frac{d}{dx}(e^{xy}) = y e^{xy} \) - \( \frac{\partial w}{\partial y} = \frac{d}{dy}(e^{xy}) = x e^{xy} \)
04
Compute Derivatives of x and y with respect to t
Now, compute the derivatives of \( x \) and \( y \) with respect to \( t \): - \( \frac{d x}{d t} = \frac{d}{d t}\left(\frac{4}{2t+1}\right) = -\frac{8}{(2t+1)^2} \) - \( \frac{d y}{d t} = \frac{d}{d t}(3t+5) = 3 \)
05
Evaluate and Substitute at t=0
Evaluate each component at \( t = 0 \): - \( x = \frac{4}{2(0)+1} = 4 \) - \( y = 3(0) + 5 = 5 \) Substitute these values into the original expressions: - \( \frac{\partial w}{\partial x} = 5 e^{20} \) - \( \frac{\partial w}{\partial y} = 4 e^{20} \) - \( \frac{d x}{d t} = -\frac{8}{1^2} = -8 \) - \( \frac{d y}{d t} = 3 \)
06
Final computation of the derivative
Finally, substitute everything into the chain rule formula: \[ \frac{d w}{d t} = (5 e^{20})(-8) + (4 e^{20})(3) = -40 e^{20} + 12 e^{20} \] Combine and simplify: \[ \frac{d w}{d t} = -28 e^{20} \] This gives us the rate of change of \( w \) with respect to \( t \) at \( t = 0 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
Partial derivatives are a fundamental concept when dealing with functions of multiple variables. In this exercise, the function is given as \( w = e^{xy} \). Since \( w \) depends on both variables \( x \) and \( y \), we need to find the effect of each variable separately on \( w \) while keeping the other variable constant.
Here is how you do it:
Here is how you do it:
- To find \( \frac{\partial w}{\partial x} \), consider \( y \) as a constant and differentiate \( e^{xy} \) with respect to \( x \). This gives \( y e^{xy} \).
- Similarly, to find \( \frac{\partial w}{\partial y} \), treat \( x \) as a constant while differentiating \( e^{xy} \) in terms of \( y \), resulting in \( x e^{xy} \).
Derivatives
Derivatives often represent the rate of change of one quantity with respect to another. In our problem, we need to find \( \frac{dw}{dt} \), which involves taking derivatives of functions of \( t \). Here, both \( x \) and \( y \) are given as functions of \( t \):
- The derivative of \( x = \frac{4}{2t+1} \) with respect to \( t \) is \( -\frac{8}{(2t+1)^2} \). This derivative tells us how \( x \) changes when \( t \) changes.
- For \( y = 3t + 5 \), its derivative with respect to \( t \) is simply \( 3 \), representing the rate at which \( y \) changes as \( t \) changes.
Function Composition
Function composition involves creating a function from two or more functions, where the output of one function becomes the input of another. In this exercise, you can think of \( w = e^{xy} \) as being composed from the separate functions \( x(t) \) and \( y(t) \). Therefore, \( w \) is indirectly a function of \( t \).
To find the derivative \( \frac{dw}{dt} \), apply the chain rule for multivariable functions:
To find the derivative \( \frac{dw}{dt} \), apply the chain rule for multivariable functions:
- First, take the partial derivatives of \( w \) with respect to \( x \) and \( y \), which are \( y e^{xy} \) and \( x e^{xy} \), respectively.
- Next, compute the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \), which ensure that we account for how \( x \) and \( y \) themselves are changing with \( t \).
- Combine these with the chain rule formula: \( \frac{dw}{dt} = \frac{\partial w}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial w}{\partial y} \cdot \frac{dy}{dt} \).
