Question

Please helpppp me with calculus

Answer 1

The definition says

$f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h$

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With $f(x)=(x+2)^2+8$, we get

$f'(x)=\displaystyle\lim_{h\to0}\frac{((x+h+2)^2+8)-((x+2)^2+8)}h=\lim_{h\to0}\frac{(x+2)^2+2h(x+2)+h^2+8-(x+2)^2-8}h$

$f'(x)=\displaystyle\lim_{h\to0}\frac{2h(x+2)+h^2}h=\lim_{h\to0}2(x+2)+h=2(x+2)$

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With $f(x)=\sqrt{2x-5}$:

$f'(x)=\displaystyle\lim_{h\to0}\frac{\sqrt{2(x+h)-5}-\sqrt{2x-5}}h=\lim_{h\to0}\frac{(2(x+h)-5)-(2x-5)}{h(\sqrt{2(x+h)-5}+\sqrt{2x-5})}$

$f'(x)=\displaystyle\lim_{h\to0}\frac{2h}{h(\sqrt{2x+2h-5}+\sqrt{2x-5})}=\lim_{h\to0}\frac2{\sqrt{2x+2h-5}+\sqrt{2x-5}}$

$f'(x)=\dfrac2{\sqrt{2x-5}+\sqrt{2x-5}}=\dfrac1{\sqrt{2x-5}}$

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With $f(x)=\dfrac1{3x-1}$:

$f'(x)=\displaystyle\lim_{h\to0}\frac{\frac1{3(x+h)-1}-\frac1{3x-1}}h=\lim_{h\to0}\frac{(3x-1)-(3(x+h)-1)}{h(3(x+h)-1)(3x-1)}$

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

$f'(x)=-\dfrac3{(3x-1)^2}$