Revision as of 08:57, 22 December 2020

Derivatives

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 ${\frac {1}{y}}\cdot {\frac {dy}{dx}}=\ln x+1$ .
Hence ${\frac {dy}{dx}}=y(\ln x+1)=x^{x}(\ln x+1)$ .

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 = Derivatives =
+== The derivative of <math>\frac{d}{dx} x^x</math> ==
+* Let <math>y = x^x</math>.
+* Take <math>\ln</math> of both sides: <math>\ln y = x \ln x</math>.
+* Differentiate both sides: <math>\frac{d}{dx} \ln y = \frac{d}{dx} x \ln x</math>.
+* Apply the chain rule on the left-hand side: <math>\frac{d}{dx} \ln y = \frac{1}{y} \cdot \frac{dy}{dx}</math>.
+* Apply the product rule on the right-hand side: <math>\frac{d}{dx} x \ln x = 1 \cdot \ln x + x \cdot \frac{1}{x} = \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>.
 = Integrals =
 = Limits =