Question

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

Answer 1

`f'(x)= \lim_(h \to 0) (f(x+h)-f(x))/(h) \\\\-----------------\\f(x)=4x^2-4x-3\\\\f'(x)= \lim_(h \to 0) (4(x+h)^2-4(x+h)-3-(4x^2-4x-3))/(h)=\\\\=\lim_(h \to 0) (4(x^2+2xh+h^2)-4x-4h-3-4x^2+4x+3)/(h)=\\\\=\lim_(h \to 0) (4x^2+8xh+4h^2-4x-4h-3-4x^2+4x+3)/(h)=\\\\=\lim_(h \to 0) (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)(f(x+h)-f(x))/(h)\\ f'(x)=\lim_(h\to0)(4(x+h)^2-4(x+h)-3-(4x^2-4x-3))/(h)\\ f'(x)=\lim_(h\to0)(4(x^2+2hx+h^2)-4x-4h-3-4x^2+4x+3)/(h)\\ f'(x)=\lim_(h\to0)(4x^2+8hx+4h^2-4h-4x^2)/(h)\\ f'(x)=\lim_(h\to0)(8hx+4h^2-4h)/(h)\\ f'(x)=\lim_(h\to0){8x+4h-4}\\ f'(x)=8x+4\cdot0-4\\ f'(x)=8x-4\\`