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Calculus Archive: Questions from January 10, 2023

• topology
  2. Sean \( X=(0,1) \) y \( \tau=\{(1 / n, 1) \mid n \in \mathbb{N}, n \geq 2\} \cup\{\emptyset, X\} \). Determine si la función \( f:(X, \tau) \longrightarrow \mathbb{R} \) \[ f(x)=\sin (\pi / 2 x) \
  2 answers
• topology
  5. Determine si las siguientes funciones de \( \mathbb{R} \) en \( \mathbb{R} \) son continuas. \[ f(x)=\left\{\begin{array}{ll} -1, & x \leq 0 \\ 1, & x>0 \end{array}\right. \] \[ g(x)=\left\{\begin{
  2 answers

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  Q3: The Chain rule. (20 points) a) \( y=\left(2 x^{5}+6\right)^{2} \) b) \( y=\sqrt[3]{3-8 x} \) c) \( y=\frac{5}{(2 x-3)^{2}} \) d) \( y=\left(5 x^{3}-4 x^{2}+3 x+8\right)^{-2} \)
  2 answers

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  Q3: The Chain rule. (20 points) a) \( y=\left(2 x^{5}+6\right)^{2} \) b) \( y=\sqrt[3]{3-8 x} \) c) \( y=\frac{5}{(2 x-3)^{2}} \) d) \( y=\left(5 x^{3}-4 x^{2}+3 x+8\right)^{-2} \)
  1 answer

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  Q3: The Chain rule. (20 points) a) \( y=\left(2 x^{5}+6\right)^{2} \) b) \( y=\sqrt[3]{3-8 x} \) c) \( y=\frac{5}{(2 x-3)^{2}} \) d) \( y=\left(5 x^{3}-4 x^{2}+3 x+8\right)^{-2} \)
  2 answers
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  Q3: The Chain rule. ( 20 points) a) \( y=\left(2 x^{5}+6\right)^{2} \) b) \( y=\sqrt[3]{3-8 x} \) c) \( y=\frac{5}{(2 x-3)^{2}} \) d) \( y=\left(5 x^{3}-4 x^{2}+3 x+8\right)^{-2} \)
  2 answers
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  \( 17-18 \) Find \( y^{\prime}(1) \) 17. \( y=5 x^{2}-3 x+1 \) 18. \( y=\frac{x^{3 / 2}+2}{x} \)
  2 answers

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  f \( y=x^{x}, x>0 \), then \( y^{\prime}=? \) (a) \( x(1+\ln x) \quad \) lny \( =x^{\prime \prime} \ln x \) (b) \( x^{x}\left(1+\frac{1}{x}\right) \) c) \( x\left(1+\frac{1}{x}\right) \) 1) \( x^{2}\l
  2 answers