Question

Buna ziua, va rog frumos ajutați-ma la exercitiu 18!

Answer 1

$\displaystyle \mathtt{18.~f:(0;+\infty)\rightarrow\mathbb{R},~f(x)=(x-3) \sqrt{x} } \\ \\ \mathtt{f'(x)=\left[(x-3) \sqrt{x} \right]'=(x-3)'\cdot \sqrt{x} +(x-3)\cdot\left( \sqrt{x} \right)'=}\\ \\ \mathtt{=(x'-3')\cdot \sqrt{x} +(x-3)\cdot \frac{1}{2 \sqrt{x} }=(1-0) \cdot \sqrt{x} + \frac{x-3}{2 \sqrt{x} } =}\\ \\ \mathtt{= \sqrt{x} + \frac{x-3}{2 \sqrt{x} }= \frac{2x+x-3}{2 \sqrt{x} }= \frac{3x-3}{2 \sqrt{x} } }$

$\displaystyle \mathtt{19.~f:\mathbb{R}\rightarrow\mathbb{R},~f(x)=3^x-\left( \frac{1}{2}\right)^x }\\ \\ \mathtt{f'(x)=\left[3^x-\left( \frac{1}{2}\right)^x\right]'=\left(3^x\right)'-\left[\left( \frac{1}{2}\right)^x\right]'=3^xln~3-\left( \frac{1}{2}\right)^xln~ \frac{1}{2}=}\\ \\ \mathtt{=3^xln~3+\left( \frac{1}{2}\right)^xln~2 }$