Table of contents
- 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
3. Techniques of Differentiation
The Chain Rule
Problem 89
Textbook Question
Second derivatives Find d²y/dx² for the following functions.
y = e^-2x²
![](/channels/images/assetPage/verifiedSolution.png)
1
Step 1: Identify the function y = e^{-2x^2}. This is an exponential function where the exponent is a function of x, specifically -2x^2.
Step 2: Find the first derivative, dy/dx. Use the chain rule, which states that if y = e^u, then dy/dx = e^u * du/dx. Here, u = -2x^2, so du/dx = -4x.
Step 3: Apply the chain rule to find dy/dx = e^{-2x^2} * (-4x) = -4x * e^{-2x^2}.
Step 4: Find the second derivative, d²y/dx². Differentiate dy/dx = -4x * e^{-2x^2} using the product rule, which states that if u and v are functions of x, then d(uv)/dx = u'v + uv'.
Step 5: Apply the product rule: Let u = -4x and v = e^{-2x^2}. Then, u' = -4 and v' = e^{-2x^2} * (-4x) (using the chain rule again). Substitute these into the product rule formula to find d²y/dx².
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