## Calculus: Early Transcendentals (2nd Edition)

$a.$ $f'(x)=2ax+b$ $b.$ $f'(x)=8x-3$ $c.$ $f'(1)=5.$
$a.$ By the definition of the derivative we have $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}= \lim_{h\to0}\frac{a(x+h)^2+b(x+h)+c-(ax^2+bx+c)}{h}=\lim_{h\to0}\frac{a(x^2+h^2+2hx)+bx+bh+c-ax^2-bx-c}{h}=\lim_{h\to0}\frac{ax^2+ah^2+2ahx+bx+bh+c-ax^2-bx-c}{h}=\lim_{h\to0}\frac{ah^2+2ahx+bh}{h}=\lim_{h\to0}(ah+2ax+b)=a\cdot0+2ax+b=2ax+b.$$ $b.$ We have shown that the derivative of any function of the form of $ax^2+bx+c$ is equal to $2ax+b$ at every point $x$. If we apply that to the given function we get $$f'(x)=2\cdot4x-3=8x-3.$$ $c.$ We just have to put $x=1$ into the previous expression $$f'(1) = 8\cdot1-3=5.$$