Difference between revisions of "Mathematics/Calculus/Corner cases"

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(Added a corner case.)
(Edited a corner case: added another solution.)
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== The derivative of <math>\frac{d}{dx} x^x</math> ==
== The derivative of <math>\frac{d}{dx} x^x</math> ==
=== Question ===
What is <math>\frac{d}{dx} x^x</math>?
=== Solution 1 ===


* Let <math>y = x^x</math>.
* Let <math>y = x^x</math>.
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* Putting it together, we have <math>\frac{1}{y} \cdot \frac{dy}{dx} = \ln x + 1</math>.
* Putting it together, we have <math>\frac{1}{y} \cdot \frac{dy}{dx} = \ln x + 1</math>.
* Hence <math>\frac{dy}{dx} = y (\ln x + 1) = x^x (\ln x + 1)</math>.
* Hence <math>\frac{dy}{dx} = y (\ln x + 1) = x^x (\ln x + 1)</math>.
=== Solution 2 ===
* Note that <math>x = e^{\ln x}</math>, so <math>x^x = (e^{\ln x})^x = e^{x \ln x} </math>.
* Applying the chain rule, <math>\frac{d}{dx} x^x = \frac{d}{dx} e^{x \ln x} = e^{x \ln x} \frac{d}{dx} x \ln x</math>.
* Applying the product rule, <math>\frac{d}{dx} x \ln x = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1</math>.
* Therefore <math>\frac{d}{dx} x^x = e^{x \ln x} (\ln x + 1) = x^x (\ln x + 1)</math>.


= Integrals =
= Integrals =


= Limits =
= Limits =

Revision as of 09:06, 22 December 2020

Derivatives

The derivative of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx} x^x}

Question

What is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx} x^x} ?

Solution 1

  • Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = x^x} .
  • Take Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ln} of both sides: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ln y = x \ln x} .
  • Differentiate both sides: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx} \ln y = \frac{d}{dx} x \ln x} .
  • Apply the chain rule on the left-hand side: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx} \ln y = \frac{1}{y} \cdot \frac{dy}{dx}} .
  • Apply the product rule on the right-hand side: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx} x \ln x = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1} .
  • Putting it together, we have Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{y} \cdot \frac{dy}{dx} = \ln x + 1} .
  • Hence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dx} = y (\ln x + 1) = x^x (\ln x + 1)} .

Solution 2

  • Note that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x = e^{\ln x}} , so Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^x = (e^{\ln x})^x = e^{x \ln x} } .
  • Applying the chain rule, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx} x^x = \frac{d}{dx} e^{x \ln x} = e^{x \ln x} \frac{d}{dx} x \ln x} .
  • Applying the product rule, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx} x \ln x = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1} .
  • Therefore Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx} x^x = e^{x \ln x} (\ln x + 1) = x^x (\ln x + 1)} .

Integrals

Limits

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