- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 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 Integrals3h 25m
3. Techniques of Differentiation
The Chain Rule
Problem 59
Textbook Question
Find the value of dy/dt at t = 0 if y = 3 sin 2x and x = t² + π.

1
First, identify the given functions: y = 3 \sin(2x) and x = t^2 + \pi. We need to find \frac{dy}{dt} at t = 0.
Use the chain rule to find \frac{dy}{dt}. The chain rule states that \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}.
Differentiate y = 3 \sin(2x) with respect to x to find \frac{dy}{dx}. Using the chain rule again, \frac{dy}{dx} = 3 \cos(2x) \cdot 2 = 6 \cos(2x).
Differentiate x = t^2 + \pi with respect to t to find \frac{dx}{dt}. The derivative is \frac{dx}{dt} = 2t.
Substitute \frac{dy}{dx} and \frac{dx}{dt} into the chain rule formula: \frac{dy}{dt} = 6 \cos(2x) \cdot 2t. Evaluate this expression at t = 0, remembering that x = t^2 + \pi, so at t = 0, x = \pi.
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