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 75
Textbook Question
27–76. Calculate the derivative of the following functions.
y = √f(x), where f is differentiable and nonnegative at x.
![](/channels/images/assetPage/verifiedSolution.png)
1
Step 1: Recognize that the function y = \sqrt{f(x)} can be rewritten using exponent notation as y = [f(x)]^{1/2}. This will make it easier to apply the chain rule.
Step 2: Apply the chain rule for differentiation. The chain rule states that if you have a composite function y = g(h(x)), then the derivative y' is g'(h(x)) * h'(x).
Step 3: Differentiate the outer function g(u) = u^{1/2} with respect to u. The derivative is g'(u) = \frac{1}{2}u^{-1/2}.
Step 4: Substitute back the inner function f(x) into the derivative of the outer function. This gives us \frac{1}{2}[f(x)]^{-1/2}.
Step 5: Multiply the result from Step 4 by the derivative of the inner function f(x), which is f'(x). Therefore, the derivative of y = \sqrt{f(x)} is \frac{1}{2}[f(x)]^{-1/2} \cdot f'(x).
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