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Second partial derivatives Find the four second partial derivatives of the following functions. $$f(x, y)=\sin ^{2}\left(x^{3} y\right)$$

Short Answer

Expert verified
Question: Find the second partial derivatives of the function $$f(x, y) = \sin^2(x^3y)$$. Answer: The second partial derivatives of the given function are: $$f_{xx}(x, y) = 18x^4y^2\cos^2(x^3y) - 6x^2y\sin^2(x^3y) -6x^2\sin(2{x^3y})$$ $$f_{xy}(x, y) = 6x^5y\cos^2(x^3y) - 3x^3\sin^2(x^3y) - 3x^2y\sin(2{x^3y})$$ $$f_{yx}(x, y) = 6x^5y\cos^2(x^3y) - 3x^3\sin^2(x^3y) - 3x^2y\sin(2{x^3y})$$ $$f_{yy}(x, y) = 2x^6\cos^2(x^3y) - 2x^6\sin^2(x^3y) - 6x^3\sin(2{x^3y})$$

Step by step solution

01

Find the first order partial derivatives

To find the first partial derivative of $$f(x, y) = \sin^2(x^3y)$$, we will compute $$f_x(x, y)$$ with respect to x and $$f_y(x, y)$$ with respect to y. Using the chain rule, we have: $$f_x(x, y) = 2\sin(x^3y)\cos(x^3y)(3x^2y)$$ and $$f_y(x, y) = 2\sin(x^3y)\cos(x^3y)(x^3)$$
02

Find the second order partial derivatives

Now, we will calculate the second order partial derivatives using the first order partial derivatives found in the previous step. For the second order partial derivative with respect to x (twice), we have: $$f_{xx}(x, y) = \frac{\partial}{\partial x}\left(2\sin(x^3y)\cos(x^3y)(3x^2y)\right)$$ For the mixed second order partial derivative with respect to x and y, we have: $$f_{xy}(x, y) = \frac{\partial}{\partial y}\left(2\sin(x^3y)\cos(x^3y)(3x^2y)\right)$$ For the mixed second order partial derivative with respect to y and x, we have: $$f_{yx}(x, y) = \frac{\partial}{\partial x}\left(2\sin(x^3y)\cos(x^3y)(x^3)\right)$$ For the second order partial derivative with respect to y (twice), we have: $$f_{yy}(x, y) = \frac{\partial}{\partial y}\left(2\sin(x^3y)\cos(x^3y)(x^3)\right)$$
03

Obtain the four second partial derivatives

Calculate the second partial derivatives using the first order partial derivatives from Step 1: $$f_{xx}(x, y) = 18x^4y^2\cos^2(x^3y) - 6x^2y\sin^2(x^3y) -6x^2\sin(2{x^3y})$$ $$f_{xy}(x, y) = 6x^5y\cos^2(x^3y) - 3x^3\sin^2(x^3y) - 3x^2y\sin(2{x^3y})$$ $$f_{yx}(x, y) = 6x^5y\cos^2(x^3y) - 3x^3\sin^2(x^3y) - 3x^2y\sin(2{x^3y})$$ $$f_{yy}(x, y) = 2x^6\cos^2(x^3y) - 2x^6\sin^2(x^3y) - 6x^3\sin(2{x^3y})$$

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Key Concepts

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

Chain Rule
The chain rule is a fundamental concept in calculus used to differentiate composite functions. It helps us find derivatives of functions that are formed by other functions. In simpler terms, it's like peeling an onion layer by layer.

In the given exercise, our function is \( f(x, y) = ext{sin}^2(x^3y) \).This involves a composite structure where the sine function surrounds \( x^3y \).Applying the chain rule means we first differentiate the outer function, \( ext{sin}^2(x^3y)\), with respect to its inner function, \( ext{sin}(x^3y) \),and then multiply by the derivative of the inner function, \( x^3y \), with respect to either \( x \)or\( y \).

For example, when differentiating with respect to \(x\), the chain rule looks like this: 2\( ext{sin}(x^3y)\)\( ext{cos}(x^3y)\) multiplied by \(3x^2y\)— the derivative of the inner function with respect to \(x\).Understanding the chain rule helps break down complex functions like peeling off a layer one at a time.
Partial Derivatives
Partial derivatives involve taking the derivative of a multivariable function with respect to one variable, while holding the others constant. This means that we are only interested in the rate of change with respect to one particular variable.

In the context of this exercise, for the function \( f(x, y) = ext{sin}^2(x^3y) \), we find partial derivatives \( f_x(x, y) \)and \( f_y(x, y) \).
  • To find \( f_x(x, y) \), we differentiate with respect to \( x \), treating \( y \)as a constant.This gives us an insight into how the function changes as \( x \)alters while \( y \)remains fixed.
  • To find \( f_y(x, y) \), we do the opposite, differentiating with respect to \( y \)and treating \( x \)like a constant.

Partial derivatives are pivotal for understanding multivariable functions and for tasks like optimization, where we're interested in how functions vary with respect to their inputs.
Mixed Partial Derivatives
Mixed partial derivatives involve taking the partial derivative of a function with respect to multiple variables, in succession. This means we first differentiate with respect to one variable and then take this new expression and differentiate it with respect to another variable.

For our function \( f(x, y) = ext{sin}^2(x^3y) \), two mixed partial derivatives are calculated:\( f_{xy}(x, y) \)and \( f_{yx}(x, y) \).
  • The derivative \( f_{xy}(x, y) \)is obtained by first differentiating \( f_x(x, y) \)with respect to \( y \).
  • Meanwhile, \( f_{yx}(x, y) \)is calculated by taking the derivative of \( f_y(x, y) \)with respect to \( x \).

What’s fascinating about mixed partial derivatives is that under certain conditions, \( f_{xy} \) and \( f_{yx} \)are equal, which is known as Clairaut's theorem. It's important to remember this equality only holds generally when the function and its derivatives are continuous.
Trigonometric Functions
Trigonometric functions like sine and cosine are fundamental in understanding wave-like behaviors and periodic phenomena. They also play an essential role in calculus, particularly in solving and simplifying derivatives.

The given function, \( f(x, y) = ext{sin}^2(x^3y) \), effectively uses these trigonometric concepts. When differentiating trigonometric expressions:
  • The derivative of \( ext{sin}(u) \) is \( ext{cos}(u) \)times the derivative of \( u \).
  • The derivative of \( ext{cos}(u) \) is \(- ext{sin}(u)\)times the derivative of \( u \).
These fundamental derivatives are applied in the exercise using the chain rule.

Recognizing the patterns in trigonometric derivatives can simplify complex calculus problems, ensuring thorough comprehension of the function's behavior as angle measures change.

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Most popular questions from this chapter

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