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Title: Chain Rule
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Series: Real Analysis
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Chapter: Differentiable Functions
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YouTube-Title: Real Analysis 36 | Chain Rule
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Bright video: https://youtu.be/g57hlenwvis
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Dark video: https://youtu.be/gosLZFWrkf0
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Ad-free video: Watch Vimeo video
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Quiz: Test your knowledge
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Dark-PDF: Download PDF version of the dark video
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Print-PDF: Download printable PDF version
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Exercise Download PDF sheets
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Thumbnail (bright): Download PNG
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Thumbnail (dark): Download PNG
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Subtitle on GitHub: ra36_sub_eng.srt missing
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Timestamps (n/a)
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Subtitle in English (n/a)
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Quiz Content
Q1: If $f: \mathbb{R} \rightarrow \mathbb{R}$ and $g: \mathbb{R} \rightarrow \mathbb{R}$ are differentiable at $x_0$, is the function $f \circ g$ also differentiable at $x_0$?
A1: Yes!
A2: No, not in general.
Q2: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be differentiable at $x_0$ and $g: \mathbb{R} \rightarrow \mathbb{R}$ be differentiable at $f(x_0)$. Which conclusion is correct?
A1: $f \circ g$ is differentiable at $x_0$
A2: $f \circ g$ is differentiable at $f(x_0)$
A3: $g \circ f$ is differentiable at $x_0$
A4: $g \circ f$ is differentiable at $f(x_0)$
A5: $g \circ f$ is differentiable at $g(x_0)$
Q3: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be differentiable at $x_0$ and $g: \mathbb{R} \rightarrow \mathbb{R}$ be differentiable at $f(x_0)$. What is the correct formulation for the chain rule?
A1: $(f \circ g)^\prime(x_0) = f^\prime(x_0) \cdot g^\prime(x_0)$
A2: $(f \circ g)^\prime(x_0) = f^\prime(x_0) + g^\prime(x_0) + f(x_0)$
A3: $(g \circ f)^\prime(x_0) = f^\prime(x_0) \cdot g(x_0)$
A4: $(g \circ f)^\prime(x_0) = g^\prime( f(x_0) ) \cdot f^\prime(x_0)$
A5: $(g \circ f)^\prime(x_0) = g^\prime( f(x_0) ) \cdot g^\prime(x_0)$
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Last update: 2025-01
