$\bf (\stackrel{x}{9}~~,~~\stackrel{y}{-3})\impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{9^2+(-3)^2}\implies c=\sqrt{81+9}\implies c=\sqrt{90} \\\\[-0.35em] ~\dotfill$

$\bf cos(\theta )=\cfrac{\stackrel{adjacent}{9}}{\stackrel{hypotenuse}{\sqrt{90}}}\implies \stackrel{\textit{rationalizing the denominator}}{\cfrac{9}{\sqrt{90}}\cdot \cfrac{\sqrt{90}}{\sqrt{90}}\implies \cfrac{9\sqrt{90}}{90}}\implies \cfrac{\sqrt{90}}{10}\implies \cfrac{3\sqrt{10}}{10}$

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=y%20%3D%20%20%7Bx%7D%5E%7Bx%7D%20" id="TexFormula1" title="y = {x}^{x} " alt="y = {x}^{x} "

Yuki888 [10]

Answer:

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

7 0

3 years ago