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Chapter 12: Problem 7

If \(F(x, y)=5 x^{3} y^{6}-x y^{7},\) find \(\partial^{3} F(x, y) / \partial x \partial y^{2}\)

Short Answer

Expert verified
\( \frac{\partial^3 F}{\partial x \partial y^2} = 450x^2y^4 - 42y^5 \)

Step by step solution

01

Find the First Partial Derivative with respect to y

Start by differentiating the function \( F(x, y) = 5x^3y^6 - xy^7 \) with respect to \( y \). Use the power rule to get:\[ \frac{\partial F}{\partial y} = 5x^3 \cdot 6y^5 - x \cdot 7y^6 = 30x^3y^5 - 7xy^6 \]
02

Find the Second Partial Derivative with respect to y

Differentiate \( \frac{\partial F}{\partial y} = 30x^3y^5 - 7xy^6 \) with respect to \( y \) again:\[ \frac{\partial^2 F}{\partial y^2} = 30x^3 \cdot 5y^4 - 7x \cdot 6y^5 = 150x^3y^4 - 42xy^5 \]
03

Find the Third Mixed Partial Derivative with respect to x and y

Now, differentiate \( \frac{\partial^2 F}{\partial y^2} = 150x^3y^4 - 42xy^5 \) with respect to \( x \):\[ \frac{\partial^3 F}{\partial x \partial y^2} = 3\cdot 150x^2y^4 - 42y^5 = 450x^2y^4 - 42y^5 \]
04

Final Verification

Double-check each differentiation step to ensure calculations are correct. Each differentiation follows standard rules of partial differentiation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mixed Partial Derivatives
Mixed partial derivatives involve differentiating a function more than once and with respect to different variables. This process helps determine how the function changes when multiple variables are altered. For the given function, we sought the third mixed partial derivative \( \frac{\partial^3 F}{\partial x \partial y^2} \). Here:
  • The function first undergoes differentiation with respect to one variable, in this case, twice with respect to \( y \).
  • Then, the resultant expression is differentiated with respect to another variable (\( x \) here).
When finding mixed derivatives, maintaining the correct order and ensuring accuracy at each step of differentiation is crucial. This specific order reflects how each variable contributes to the behavior of the overall function.
Power Rule
The power rule is a straightforward method used in calculus for differentiating expressions of the form \( x^n \). When applying the power rule, you bring the exponent down as a coefficient and subtract one from the original exponent:
  • \( \frac{d}{dx} x^n = n \cdot x^{n-1} \)
In our exercise, you saw applications of the power rule during each differentiation step. For instance, starting from:
  • \( F(x, y) = 5x^3y^6 - xy^7 \)
The expression was broken down to simpler components:
  • \( 6y^5 \) resulted from \( y^6 \)
  • \( 5y^4 \) came from \( y^5 \) on further differentiation
Using the power rule at each stage simplifies, yet is essential in achieving accurate solutions efficiently.
Step-by-Step Solution
Breaking down complex calculus problems into smaller, manageable steps is an effective strategy to ensure accuracy. Here's how each stage simplifies the overall problem:
  • Step 1: Differentiate the original function with respect to \( y \) to obtain the first partial derivative.
  • Step 2: Differentiate again with respect to \( y \) to find the second partial derivative.
  • Step 3: Change focus and differentiate with respect to \( x \) to achieve the mixed partial derivative.
Each step involves applying known rules and converting complex expressions into simpler derivatives. Additionally, verifying each step increases confidence in the obtained result. For students, following a methodical approach not only leads to the correct solution but also builds proficiency in handling advanced calculus operations.

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Most popular questions from this chapter

If the temperature of a plate at the point \((x, y)\) is \(T(x, y)=10+x^{2}-y^{2},\) find the path a heat-seeking particle (which always moves in the direction of greatest increase in temperature) would follow if it starts at (-2,1) . Hint: The particle moves in the direction of the gradient $$ \nabla T=2 x \mathbf{i}-2 y \mathbf{j} $$ We may write the path in parametric form as $$ \mathbf{r}(t)=x(t) \mathbf{i}+y(t) \mathbf{j} $$ and we want \(x(0)=-2\) and \(y(0)=1 .\) To move in the required direction means that \(\mathbf{r}^{\prime}(t)\) should be parallel to \(\nabla T\). This will be satisfied if $$ \frac{x^{\prime}(t)}{2 x(t)}=-\frac{y^{\prime}(t)}{2 y(t)} $$ together with the conditions \(x(0)=-2\) and \(y(0)=1\). Now solve this differential equation and evaluate the arbitrary constant of inteoration

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Find the maximum and minimum values of \(f(x, y)=10+x+y\) on the disk \(x^{2}+y^{2} \leq 9 .\) Hint: Parametrize the boundary by \(x=3 \cos t, y=3 \sin t, 0 \leq t \leq 2 \pi\).
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