Problem 17 - APEX Calculus

Question

Find \(f_{x}, f_{y}, f_{x x}, f_{y y}, f_{x y}\) and \(f_{y x}\). $$ f(x, y)=\cos \left(5 x y^{3}\right) $$

Step-by-Step Solution

Answer

\(f_x = -5y^3 \sin(5xy^3)\), \(f_y = -15xy^2 \sin(5xy^3)\), \(f_{xx} = -25y^6 \cos(5xy^3)\), \(f_{yy} = -30xy\sin(5xy^3) - 225x^2y^4 \cos(5xy^3)\), \(f_{xy} = f_{yx} = -15y^2 \sin(5xy^3) - 75xy^5 \cos(5xy^3)\).

Step 1: Find the first partial derivative with respect to x

To find \(f_x\), differentiate \(f(x, y) = \cos(5xy^3)\) with respect to \(x\), treating \(y\) as a constant.\[f_x = \frac{d}{dx}\left( \cos(5xy^3) \right) = -\sin(5xy^3) \times \frac{d}{dx}(5xy^3) \] The derivative of \(5xy^3\) with respect to \(x\) is \(5y^3\). Thus, \[f_x = -\sin(5xy^3) \cdot 5y^3 = -5y^3 \sin(5xy^3)\]

Step 2: Find the first partial derivative with respect to y

Now, find \(f_y\) by differentiating \(f(x, y) = \cos(5xy^3)\) with respect to \(y\), treating \(x\) as a constant.\[f_y = \frac{d}{dy} \left(\cos(5xy^3)\right) = -\sin(5xy^3) \times \frac{d}{dy}(5xy^3) \] The derivative of \(5xy^3\) with respect to \(y\) is \(15xy^2\). Thus, \[f_y = -\sin(5xy^3) \cdot 15xy^2 = -15xy^2 \sin(5xy^3) \]

Step 3: Find the second partial derivative with respect to x twice

Determine \(f_{xx}\) by differentiating \(f_x = -5y^3 \sin(5xy^3)\) with respect to \(x\).\[f_{xx} = \frac{d}{dx} \left(-5y^3 \sin(5xy^3)\right) = -5y^3 \cos(5xy^3) \cdot \frac{d}{dx}(5xy^3)\] Substitute \(\frac{d}{dx}(5xy^3) = 5y^3\)\[f_{xx} = -5y^3 \cdot 5y^3 \cdot \cos(5xy^3) = -25y^6 \cos(5xy^3)\]

Step 4: Find the second partial derivative with respect to y twice

Find \(f_{yy}\) by differentiating \(f_y = -15xy^2 \sin(5xy^3)\) with respect to \(y\).\[f_{yy} = \frac{d}{dy} \left(-15xy^2 \sin(5xy^3)\right) = -15x\left(2y\sin(5xy^3) + y^2\cos(5xy^3) \cdot \frac{d}{dy}(5xy^3)\right)\] Where \(\frac{d}{dy}(5xy^3) = 15xy^2\), \[f_{yy} = -30xy\sin(5xy^3) - 225x^2y^4 \cos(5xy^3)\]

Step 5: Find the mixed partial derivative with respect to x and then y

To find \(f_{xy}\), differentiate \(f_x = -5y^3 \sin(5xy^3)\) with respect to \(y\).\[f_{xy} = \frac{d}{dy}\left(-5y^3 \sin(5xy^3)\right) = -5(3y^2 \sin(5xy^3) + y^3 \cos(5xy^3) \cdot 15xy^2)\] \[f_{xy} = -15y^2 \sin(5xy^3) - 75xy^5 \cos(5xy^3)\]

Step 6: Find the mixed partial derivative with respect to y and then x

For \(f_{yx}\), differentiate \(f_y = -15xy^2 \sin(5xy^3)\) with respect to \(x\).\[f_{yx} = \frac{d}{dx} \left(-15xy^2 \sin(5xy^3)\right) = -15y^2\left(\sin(5xy^3) + x \cos(5xy^3) \cdot 5y^3\right)\] \[f_{yx} = -15y^2 \sin(5xy^3) - 75xy^5 \cos(5xy^3)\]

Key Concepts

Mixed Partial Derivatives

When working with functions of multiple variables, understanding mixed partial derivatives is crucial. A mixed partial derivative is where we differentiate a function with respect to two different variables in sequence. In this exercise, we came across

• \(f_{xy}\): the derivative first with respect to \(x\), then \(y\)
• \(f_{yx}\): the derivative first with respect to \(y\), then \(x\)

These derivatives are significant in checking the interchangeability of differentiation operations. According to Clairaut’s Theorem, for continuous functions and continuous partial derivatives, \(f_{xy} = f_{yx}\).
In the provided solution, we confirmed that both mixed partial derivatives equate to \(-15y^2 \sin(5xy^3) - 75xy^5 \cos(5xy^3)\), thus demonstrating the theorem. Paying close attention to these calculations builds a deeper understanding of how partial derivatives work in calculus.

Second Partial Derivatives

Second partial derivatives provide insights into a function's concavity and curvature. This exercise involves calculating \(f_{xx}\) and \(f_{yy}\), for which we differentiate the first partial derivatives again by the same variable.

• \(f_{xx}\): The second partial derivative of \(f(x,y)\) twice concerning \(x\).
• \(f_{yy}\): The second partial derivative of \(f(x,y)\) twice concerning \(y\).

These derivatives help analyze how the slope of a curve changes at different points. In our example, we calculated \(f_{xx} = -25y^6 \cos(5xy^3)\) and \(f_{yy} = -30xy \sin(5xy^3) - 225x^2y^4 \cos(5xy^3)\). Understanding these provides useful insights into the behavior of the involved function, particularly when examining surfaces or optimization problems.

Trigonometric Functions

Trigonometric functions often appear in calculus, and understanding how to differentiate them is essential. In our problem, the function was \(f(x, y) = \cos(5xy^3)\). When differentiating trigonometric functions, remember:

• For differentiation: \(\frac{d}{dx}(\cos(u)) = -\sin(u) \cdot \frac{d}{dx}(u)\)
• For integration: \(\int \sin(u) \, du = -\cos(u) + C\)

Utilizing these rules can significantly simplify the process. We applied them to derive \(f_x\), \(f_y\), and the second derivatives. Furthermore, understanding the trigonometric identities can aid in simplifying expressions, optimization, and solving equations. Through these exercises, gaining comfort with differentiating trigonometric parts of the function empowers students when tackling more complex calculus problems.