Table of contents

0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

2. Intro to Derivatives

Derivatives as Functions

Calculus

2. Intro to Derivatives

Derivatives as Functions

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Multiple Choice

Given $f (x, y) = \sin^{2} (m x + n y)$ , which of the following is the correct expression for the second mixed partial derivative $\frac{\partial^{2} f}{\partial x \partial y}$ ?

m n \cos (m x + n y)

2 m n \sin (2 (m x + n y))

m^{2} n^{2} \sin^{2} (m x + n y)

2 m n \cos (2 (m x + n y))

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Verified step by step guidance

Step 1: Begin by understanding the problem. You are tasked with finding the second mixed partial derivative $ \frac{\partial^2 f}{\partial x \partial y} $ for the given function $ f(x, y) = \sin^2(mx + ny) $. This involves differentiating the function first with respect to $ x $, and then with respect to $ y $.

Step 2: Compute the first partial derivative of $ f(x, y) $ with respect to $ x $. Use the chain rule for differentiation. The derivative of $ \sin^2(u) $ is $ 2\sin(u)\cos(u) $, and the derivative of $ u = mx + ny $ with respect to $ x $ is $ m $. Thus, $ \frac{\partial f}{\partial x} = 2\sin(mx + ny)\cos(mx + ny) \cdot m $.

Step 3: Simplify the expression for $ \frac{\partial f}{\partial x} $. Using the trigonometric identity $ \sin(2u) = 2\sin(u)\cos(u) $, rewrite $ \frac{\partial f}{\partial x} $ as $ m\sin(2(mx + ny)) $.

Step 4: Compute the second mixed partial derivative $ \frac{\partial^2 f}{\partial x \partial y} $ by differentiating $ \frac{\partial f}{\partial x} $ with respect to $ y $. Again, apply the chain rule. The derivative of $ \sin(2u) $ is $ 2\cos(2u) $, and the derivative of $ u = mx + ny $ with respect to $ y $ is $ n $. Thus, $ \frac{\partial^2 f}{\partial x \partial y} = m \cdot 2\cos(2(mx + ny)) \cdot n $.

Step 5: Simplify the final expression. Combine constants $ m $ and $ n $ with $ 2 $ to get $ \frac{\partial^2 f}{\partial x \partial y} = 2mn\cos(2(mx + ny)) $. This matches the correct answer provided in the problem.