Please helpppp me with calculus

Question 1

Question 2

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)=√(2x-5) :

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

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

$f'(x)=\frac2{√(2x-5)+√(2x-5)}=\frac1{√(2x-5)}$

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With
$f(x)=\frac1{3x-1}$ :

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

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

$f'(x)=-\frac3{(3x-1)^2}$

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