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Title: Derivatives of Inverse Functions
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Series: Real Analysis
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Chapter: Differentiable Functions
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YouTube-Title: Real Analysis 39 | Derivatives of Inverse Functions
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Bright video: https://youtu.be/h0nBAMhdSMk
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Dark video: https://youtu.be/0cktpu5Hxjk
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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: ra39_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: What is not a property of the exponential function $\exp: \mathbb{R} \rightarrow \mathbb{R}$?
A1: $\exp$ is continuous.
A2: $\exp(x+y) = \exp(x) \exp(y)$
A3: $\exp$ is strictly monotonically increasing.
A4: $\exp(2) = e^2$
A5: $\exp: \mathbb{R} \rightarrow [0,\infty)$ is bijective.
Q2: The logarithm function $\log: (0, \infty) \rightarrow \mathbb{R}$ is defined as the inverse function of $\exp: \mathbb{R} \rightarrow (0,\infty)$. Is $\log$ differentiable?
A1: No, nowhere!
A2: Yes, but only for $x_0 =1$.
A3: Yes, everywhere.
Q3: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be invertible and differentiable at $x_0$. What is the correct inversion formula if $f^{-1}$ is also differentiable at $y_0 = f(x_0)$?
A1: $ \displaystyle (f^{-1})^\prime(y_0) = f^\prime(x_0)$
A2: $ \displaystyle (f^{-1})^\prime(y_0) = \frac{1}{f^\prime(x_0)}$
A3: $ \displaystyle (f^{-1})^\prime(y_0) = \frac{1}{f^\prime(f^{-1}(x_0))}$
A4: $ \displaystyle (f^{-1})^\prime(y_0) = (f^\prime(f^{-1}(x_0)))^{-1}$
Q4: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be invertible and differentiable at $x_0$. What is the correct assumption to conclude that $f^{-1}$ is also differentiable at $y_0 = f(x_0)$?
A1: $f^{-1}$ is invertible at $y_0$
A2: $f^\prime(x_0) \neq 0$
A3: $f^\prime(x_0) \neq 0$ and $f^{-1}$ is continuous at $y_0$
A4: $f^\prime(x_0) \neq 0$ or $f^{-1}$ is continuous at $y_0$
A5: $f^\prime(x_0) = 0$ or $f^{-1}$ is continuous at $y_0$
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Last update: 2025-01
