Hi Yura_chan,
I've done the algebra below but you reall want to graph this and try to understand what is happening :/
f(X) = 2x^2 + 4 at x=1 Find the derivative, both a and b should have the same answer.
a.) Use equation lim as x -> c f(x)-f(c)/x-c
\(\displaystyle \lim_{x\rightarrow1} \;\;\frac{f(x)-f(1)}{x-1}\\ =\displaystyle \lim_{x\rightarrow1} \;\;\frac{2x^2+4-6}{x-1}\\ =\displaystyle \lim_{x\rightarrow1} \;\;\frac{2x^2-2}{x-1}\\ =\displaystyle \lim_{x\rightarrow1} \;\;\frac{2(x^2-1)}{x-1}\\ =\displaystyle \lim_{x\rightarrow1} \;\;\frac{2(x-1)(x+1)}{x-1}\\ =\displaystyle \lim_{x\rightarrow1} \;\;\frac{2(x+1)}{1}\\ =2*(1+1)\\ =4\)
b.) Use equation lim as h ->0 f(c+h)-f(c)/h
\(\displaystyle \lim_{h\rightarrow0} \;\;\frac{f(c+h)-f(c)}{c+h-c}\\ =\displaystyle \lim_{h\rightarrow0} \;\;\frac{f(1+h)-f(1)}{h}\\ =\displaystyle \lim_{h\rightarrow0} \;\;\frac{2h^2+4h+6-6}{h}\\ =\displaystyle \lim_{h\rightarrow0} \;\;\frac{2h^2+4h}{h}\\ =\displaystyle \lim_{h\rightarrow0} \;\;\frac{2h(h+2)}{h}\\ =\displaystyle \lim_{h\rightarrow0} \;\;\frac{2(h+2)}{1}\\ =2*(0+2)\\ =4\)
this means that the gradient of the tangent to the curve f(x)=2x^2+4 is 4 when x=1
+118728 
