Using first principle: f'(x)=lim (h→0) [(x+h)² - x²]/h = lim (h→0) [2xh + h²]/h = 2x + h. As h→0, derivative = 2x.
f'(x)=lim (h→0) [(x+h)³ - x³]/h = lim (h→0) [3x²h+3xh²+h³]/h = 3x²+3xh+h². As h→0, derivative = 3x².
f'(x)=lim (h→0) [(1/(x+h)) - (1/x)]/h = lim (h→0) [-h/(x(x+h)h)] = -1/(x²).
f'(x)=lim (h→0) [(√(x+h) - √x)/h] × [(√(x+h)+√x)/(√(x+h)+√x)] = lim (h→0) [1/(√(x+h)+√x)] = 1/(2√x).
f'(x)=lim (h→0) [(x+h)-x]/h = lim (h→0) h/h = 1.
f'(x)=lim (h→0) [(c - c)/h] = 0. Derivative of a constant is zero.
f'(x)=lim (h→0) [(x+h)ⁿ - xⁿ]/h = nx^(n-1) (proved using binomial expansion).
f'(x)=lim (h→0) [(sin(x+h)-sinx)/h] = lim (h→0) [sinx(cos h - 1) + cosx sin h]/h = cos x.
f'(x)=lim (h→0) [(cos(x+h)-cosx)/h] = lim (h→0) [-cosx(1 - cos h) - sinx sin h]/h = -sin x.
f'(x)=lim (h→0) [(e^(x+h)-e^x)/h] = e^x × lim (h→0) [(e^h - 1)/h] = e^x.