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Applied Mathematics: For the Managerial, Life, and Social Sciences

Soo T. Tan

Chapter 12: Chapter 12

Problem 11

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Find the first partial derivatives of the function. $f(s, t)=\left(s^{2}-s t+t^{2}\right)^{3}$

Short Answer

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The first partial derivatives of the function $f(s, t) = (s^2 - st + t^2)^3$ are: $\frac{\partial f}{\partial s} = 3(s^2 - st + t^2)^2 \cdot (2s - t)$ $\frac{\partial f}{\partial t} = 3(s^2 - st + t^2)^2 \cdot (-s + 2t)$

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Step by step solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Write down the given function with its composite function.

The given function is $f(s, t) = (s^2 - st + t^2)^3$. We can notice that this function is a composite of two functions. We can rewrite it as: $f(s, t) = g(h(s, t))$, where $g(u) = u^3$ and $h(s, t) = s^2 - st + t^2$.

Step 2: Evaluate the derivatives of the inner and outer functions with respect to the given variables.

Now we need to find the derivatives of the inner function $h(s, t)$ and the outer function $g(u)$ with respect to our variables, $s$ and $t$. Let's first find the derivatives with respect to $s$: $\frac{\partial h}{\partial s} = 2s - t$ $\frac{\partial g}{\partial u} = 3u^2$ Now, let's find the derivatives with respect to $t$: $\frac{\partial h}{\partial t} = -s + 2t$ $\frac{\partial g}{\partial u} = 3u^2$ (the derivative of $g$ with respect to $u$ remains the same)

Step 3: Apply the chain rule to find the first partial derivatives of the given function.

Now that we have our inner and outer function derivatives, we can apply the chain rule to find our partial derivatives of the given function $f(s, t)$. First, the partial derivative with respect to $s$: $\frac{\partial f}{\partial s} = \frac{\partial g}{\partial h} \cdot \frac{\partial h}{\partial s}$ Substitute the found derivatives: $\frac{\partial f}{\partial s} = (3u^2) \cdot (2s - t)$ Now, recall that $u = h(s, t) = s^2 - st + t^2$. Substitute this back into our equation: $\frac{\partial f}{\partial s} = 3(s^2 - st + t^2)^2 \cdot (2s - t)$ Next, the partial derivative with respect to $t$: $\frac{\partial f}{\partial t} = \frac{\partial g}{\partial h} \cdot \frac{\partial h}{\partial t}$ Substitute the found derivatives: $\frac{\partial f}{\partial t} = (3u^2) \cdot (-s + 2t)$ Again, substitute $u = h(s, t) = s^2 - st + t^2$ back into our equation: $\frac{\partial f}{\partial t} = 3(s^2 - st + t^2)^2 \cdot (-s + 2t)$

Step 4: Write down the final first partial derivatives of the given function.

We found the first partial derivatives of the given function $f(s, t) = (s^2 - st + t^2)^3$: $\frac{\partial f}{\partial s} = 3(s^2 - st + t^2)^2 \cdot (2s - t)$ $\frac{\partial f}{\partial t} = 3(s^2 - st + t^2)^2 \cdot (-s + 2t)$

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Chapter 12: Chapter 12

Find the first partial derivatives of the function. \(f(s, t)=\left(s^{2}-s t+t^{2}\right)^{3}\)

Short Answer

Step by step solution

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Step 1: Write down the given function with its composite function.

Step 2: Evaluate the derivatives of the inner and outer functions with respect to the given variables.

Step 3: Apply the chain rule to find the first partial derivatives of the given function.

Step 4: Write down the final first partial derivatives of the given function.

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More chapters from the book ‘Applied Mathematics: For the Managerial, Life, and Social Sciences’

Chapter 1

Chapter 2

Chapter 3

Chapter 4

Chapter 5

Chapter 6

Chapter 7

Chapter 8

Chapter 9

Chapter 10

Chapter 11

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