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$f(x)={[2x^(2)-8x,","x in[0","6)],[sqrt(x^(2)-8)",",x >= 6]:} Dominio de la función,$

$f(x)=\left\{\begin{array}{ll} 2 x^{2}-8 x & , x \in[0,6) \\ \sqrt{x^{2}-8}, & x \geq 6 \end{array}\right.$ $\newline$ * Dominio de la función,

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Find derivatives of logarithmic functions

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Q.

f(x)=\left\{\begin{array}{ll} 2 x^{2}-8 x & , x \in[0,6) \\ \sqrt{x^{2}-8}, & x \geq 6 \end{array}\right.

\newline

* Dominio de la función,

Identify Function Form: Identify the function's form: f(x) = \left\{ \begin{array}{ll} \(2x^ $2$ - $8$ x & \text{for } x \in [ $0$ , $6$ ), \ \sqrt{x^ $2$ - $8$ } & \text{for } x \geq $6$ \end{array} \right.\}
Calculate First Part Derivative: Calculate the derivative for the first part: $\newline$ For $2x^2 - 8x$ , use the power rule. $\newline$ $d/dx(2x^2 - 8x) = 4x - 8$
Calculate Second Part Derivative: Calculate the derivative for the second part: $\newline$ For $\sqrt{x^2 - 8}$ , use the chain rule. $\newline$ Let $u = x^2 - 8$ , then the derivative of $\sqrt{u}$ is $(1/2)u^{-1/2} \frac{du}{dx}$ . $\newline$ $\frac{du}{dx} = 2x$ , $\newline$ So, $\frac{d}{dx}(\sqrt{x^2 - 8}) = (1/2)(x^2 - 8)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 - 8}}$

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