Question

Differenciate by first principles f(x) = 4x²-4x-3

Answer 1

$f'(x)= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \\\\-----------------\\f(x)=4x^2-4x-3\\\\f'(x)= \lim_{h \to 0} \frac{4(x+h)^2-4(x+h)-3-(4x^2-4x-3)}{h}=\\\\=\lim_{h \to 0} \frac{4(x^2+2xh+h^2)-4x-4h-3-4x^2+4x+3}{h}=\\\\=\lim_{h \to 0} \frac{4x^2+8xh+4h^2-4x-4h-3-4x^2+4x+3}{h}=\\\\=\lim_{h \to 0} \frac{8xh+4h^2-4h}{h}=\lim_{h \to 0} (8x+4h- 4)=8x+4\cdot0-4=8x-4\\\\f'(x)=8x-4$

Answer 2

$f(x) = 4x^2-4x-3\\ f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ f'(x)=\lim_{h\to0}\frac{4(x+h)^2-4(x+h)-3-(4x^2-4x-3)}{h}\\ f'(x)=\lim_{h\to0}\frac{4(x^2+2hx+h^2)-4x-4h-3-4x^2+4x+3}{h}\\ f'(x)=\lim_{h\to0}\frac{4x^2+8hx+4h^2-4h-4x^2}{h}\\ f'(x)=\lim_{h\to0}\frac{8hx+4h^2-4h}{h}\\ f'(x)=\lim_{h\to0}{8x+4h-4}\\ f'(x)=8x+4\cdot0-4\\ f'(x)=8x-4$