Question

alguien que me ayude de favor lo más pronto posible ​

Answer 1

Explicación paso a paso:

1. f(x)=x+3

Al aplicar la definición de derivada de una función, aplicando limites:

$\lim_{x \to0} \frac{(x+h+3)-(x+3)}{h}= \\\\ \lim_{x \to0} \frac{x+h+3-x-3}{h} =\\\\ \lim_{x \to0} \frac{h}{h}= \\\\ \lim_{x \to0} 1$

2. f(x)=3x²

Al aplicar la definición de derivada de una función, aplicando limites:

$\lim_{x \to0} \frac{3(x+h)^{2} -3x^{2} }{h}= \\\\ \lim_{x \to0} \frac{3(x^{2} +2xh+h^{2} )-3x^{2}}{h} =\\\\ \lim_{x \to0} \frac{3x^{2}+6xh+3h^{2}-3x^{2} }{h}= \\\\ \lim_{x \to0} \frac{6xh+3h^{2} }{h}= \\\\ \lim_{x \to0} \frac{h(6x+h)}{h}=\\\\ \lim_{x \to0}6x+h =\\\\ \lim (6x+h)_{x \to0}=h$

3. f(x)=$\sqrt{x-1}$

Al aplicar la definición de derivada de una función, aplicando limites:

$\lim_{x \to0} \frac{\sqrt{x+h-1}- \sqrt{x+h} }{h}= \\\\\lim_{x \to0} \frac{\sqrt{x+h-1}- \sqrt{x+h} }{h}*\frac{\sqrt{x+h-1}- \sqrt{x+h}}{\sqrt{x+h-1}- \sqrt{x+h}} =\\\\\lim_{x \to0} \frac{(\sqrt{x+h-1})^{2} - (\sqrt{x+h})^{2} }{h(\sqrt{x+h-1}- \sqrt{x+h})} =\\\\\lim_{x \to0}\frac{x+h-1-x-h}{h(\sqrt{x+h-1}- \sqrt{x+h})} =\\\\\lim_{x \to0}\frac{-1}{h(\sqrt{x+h-1}- \sqrt{x+h})} =\\\\lim(\frac{-1}{h(\sqrt{x+h-1}- \sqrt{x+h})})_{x \to0} =\frac{-1}{\sqrt{h-1}-\sqrt{h} }$

4. f(x)=mx²

Al aplicar la definición de derivada de una función, aplicando limites:

$\lim_{x \to0} \frac{m(x+h)^{2} -mx^{2} }{h}= \\\\ \lim_{x \to0} \frac{m(x^{2} +2xh+h^{2} )-mx^{2}}{h} =\\\\ \lim_{x \to0} \frac{mx^{2}+2mxh+mh^{2}-mx^{2} }{h}= \\\\ \lim_{x \to0} \frac{2mxh+mh^{2} }{h}= \\\\ \lim_{x \to0} \frac{mh(2x+h)}{h}=\\\\ \lim_{x \to0}m(2x+h) =\\\\ \lim (m(2x+h))_{x \to0}=mh$

5. f(x)=$\frac{x}{2}$

Al aplicar la definición de derivada de una función, aplicando limites:

$\lim_{x \to0} \frac{\frac{(x+h)}{2} -\frac{x}{2} }{h} =\\\\\lim_{x \to0} \frac{\frac{2(x+h)-2x}{2}}{h} =\\\\\lim_{x \to0} \frac{\frac{2x+2h-2x}{2}}{h} =\\\\\lim_{x \to0} \frac{\frac{2h}{2}}{h} =\\\\\lim_{x \to0} \frac{h}{h}=\\\\\lim_{x \to0}1$

6. f(x)=$\frac{1}{x^{2} }$

Al aplicar la definición de derivada de una función, aplicando limites:

$\lim_{x \to0} \frac{\frac{1}{(x+h )^{2}} -\frac{1}{x^{2} } }{h} =\\\\\lim_{x \to0} \frac{\frac{1}{x^{2} +2xh+h^{2}}-\frac{1}{x^{2} } }{h} =\\\\\lim_{x \to0} \frac{\frac{{x^{2} -(x^{2} +2xh+h^{2})}}{x^{2} (x^{2} +2xh+h^{2})} }{h} =\\\\\lim_{x \to0} \frac{\frac{{x^{2} -x^{2} -2xh-h^{2}}}{x^{2} (x^{2} +2xh+h^{2})} }{h}= \\\\\lim_{x \to0} \frac{\frac{{-2xh-h^{2}}}{x^{2} (x^{2} +2xh+h^{2})} }{h} =\\\\\lim_{x \to0} \frac{-2xh-h^{2} }{hx^{2} (x^{2} +2xh+h^{2}))} =$

$\lim_{x \to0}\frac{h(-2x-h)}{hx^{2}(x^{2}+2xh+h^{2} )}=\\\\\lim_{x \to0}\frac{-2x-h}{x^{2}(x^{2}+2xh+h^{2} )}=\\\\\lim(\frac{-2x-h}{x^{2}(x^{2}+2xh+h^{2} )}) _{x \to0}=\frac{-h}{0}=$

$\lim(\frac{-2x-h}{x^{2}(x^{2}+2xh+h^{2} )}) _{x \to0}=$ ∞

Listo... Espero que te sirva :)