Mathematics/Calculus/Corner cases

Derivatives

The derivative of ${\frac {d}{dx}}x^{x}$

Question

What is ${\frac {d}{dx}}x^{x}$?

Solution 1

• Let $y=x^{x}$.
• Take $\ln$ of both sides: $\ln y=x\ln x$.
• Differentiate both sides: ${\frac {d}{dx}}\ln y={\frac {d}{dx}}x\ln x$.
• Apply the chain rule on the left-hand side: ${\frac {d}{dx}}\ln y={\frac {1}{y}}\cdot {\frac {dy}{dx}}$.
• Apply the product rule on the right-hand side: ${\frac {d}{dx}}x\ln x=1\cdot \ln x+x\cdot {\frac {1}{x}}=\ln x+1$.
• 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

The integral 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 \int x^x \, dx}

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 \int x^x \, dx} ?

Solution

• We can write 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} as 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 (e^{\ln x})^x = e^{x \ln x}} .
• Consider the series expansion 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 e^{x \ln x}} :

   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 e^{x \ln x} = 1 + (x \ln x) + \frac{(x \ln x)^2}{2!} + \frac{(x \ln x)^3}{3!} + \ldots + \frac{(x \ln x)^i}{i!} + \ldots = \sum_{i=0}^{\infty} \frac{(x \ln x)^i}{i!}}
.

• We can interchange the integration and summation (we can recognize this as a special case of the Fubini/Tonelli theorems) and write

   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      \int x^x \, dx = \int \left( \sum_{i=0}^{\infty} \frac{(x \ln x)^i}{i!} \right) \, dx = \sum_{i=0}^{\infty} \left( \int \frac{(x \ln x)^i}{i!} \, dx \right) = \sum_{i=0}^{\infty} \left( \frac{1}{i!} \int x^i (\ln x)^i \, dx \right).     }

Limits

The limit 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 \lim_{x \rightarrow 0^+} 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 \lim_{x \rightarrow 0^+} x^x} ?

Solution

• 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 $x^x = (e^{\ln x})^x = e^{x \ln x}} .
• We can further rewrite this as 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^{x \ln x} = e^{\frac{\ln x}{\frac{1}{x}}}} .
• As long as 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 f} is continuous and the limit 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 g} exists at the point in question, the limit will commute with composition:

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 \lim_{x \rightarrow t} f(g(x)) = f(\lim_{x \rightarrow t} g(x)). } In our case, 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 e(\cdot)} is continuous, 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 \lim_{x \rightarrow 0^+} x^x = e^{\lim_{x \rightarrow 0^+} \frac{\ln x}{\frac{1}{x}}}. }

• The question, then, is 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 \lim_{x \tendsto 0^+} \frac{\ln x}{\frac{1}{x}}} .
• As 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 \tendsto 0^+} , 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 x \tendsto -\infty$, $\frac{1}{x} \tendsto +\infty} . In this situation we can apply l'Hôpital's 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 \lim_{x \tendsto 0^+} \frac{\ln x}{\frac{1}{x}} = \lim_{x \tendsto 0^+} \frac{\frac{d}{dx} \ln x}{\frac{d}{dx} \frac{1}{x}} = \frac{\frac{1}{x}}{-\frac{1}{x^2}} = \frac{\frac{1}{x} \cdot x^2}{-\frac{1}{x^2} \cdot x^2} = -x. }

• 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 \lim_{x \tendsto 0^+} x^x = e^0 = 1} .

The limits 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 \lim_{x \tendsto +\infty} x \sin \frac{1}{x}} and 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 \lim_{x \tendsto -\infty} x \sin \frac{1}{x}}

Question

What are 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 \lim_{x \tendsto +\infty} x \sin \frac{1}{x}} and 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 \lim_{x \tendsto -\infty} x \sin \frac{1}{x}} ?

Solution

• Let us rewrite 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 \sin \frac{1}{x}$ as $\frac{\sin \frac{1}{x}}{\frac{1}{x}}} .
• As Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \tendsto +\infty$, $\frac{1}{x} \tendsto 0} and 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 \sin \frac{1}{x} \tendsto 0} .
• 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{0}{0}} , so we can apply l'Hôpital's rule.
• Differentiating the numerator in 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{\sin \frac{1}{x}}{\frac{1}{x}}} , we obtain 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 \left(\cos \frac{1}{x}\right) \left(-\frac{1}{x^2}\right)} .
• Differentiating the denominator in 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{\sin \frac{1}{x}}{\frac{1}{x}}} , we obtain 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}{x^2}} .
• Thus

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 \lim_{x \tendsto +\infty} x \sin \frac{1}{x} = \lim_{x \tendsto +\infty} \frac{\sin \frac{1}{x}}{\frac{1}{x}} = \lim_{x \tendsto +\infty} \frac{\left(\cos \frac{1}{x}\right) \left(-\frac{1}{x^2}\right)}{-\frac{1}{x^2}} = \lim_{x \tendsto +\infty} \cos \frac{1}{x} = 1. }

• Similarly we can find 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 \lim_{x \tendsto -\infty} x \sin \frac{1}{x} = 1} .