f ( x ) = x 2 {\displaystyle f(x)=x^{2}}
d d x ( f ( x ) ) = lim h → 0 f ( x + h ) − f ( x ) h {\displaystyle {\frac {d}{dx}}(f(x))=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}}
d d x ( x 2 ) = lim h → 0 ( x + h ) 2 − x 2 h = lim h → 0 x 2 + 2 x h + h 2 − x 2 h = lim h → 0 2 x h + h 2 h = lim h → 0 2 x + h = 2 x {\displaystyle {\begin{alignedat}{4}{\frac {d}{dx}}(x^{2})\,\,&=\lim _{h\to 0}{\frac {(x+h)^{2}-x^{2}}{h}}\\&=\lim _{h\to 0}{\frac {x^{2}+2xh+h^{2}-x^{2}}{h}}\\&=\lim _{h\to 0}{\frac {2xh+h^{2}}{h}}\\&=\lim _{h\to 0}2x+h\\&=2x\end{alignedat}}}
f ( x ) = x 3 {\displaystyle f(x)=x^{3}}
d d x ( x 3 ) = lim h → 0 ( x + h ) 3 − x 3 h = lim h → 0 x 3 + 3 x 2 h + 3 x h 2 + h 3 − x 3 h = lim h → 0 3 x 2 h + 3 x h 2 + h 3 h = lim h → 0 3 x 2 + 3 x h + h 2 = 3 x 2 {\displaystyle {\begin{alignedat}{4}{\frac {d}{dx}}(x^{3})\,\,&=\lim _{h\to 0}{\frac {(x+h)^{3}-x^{3}}{h}}\\&=\lim _{h\to 0}{\frac {x^{3}+3x^{2}h+3xh^{2}+h^{3}-x^{3}}{h}}\\&=\lim _{h\to 0}{\frac {3x^{2}h+3xh^{2}+h^{3}}{h}}\\&=\lim _{h\to 0}3x^{2}+3xh+h^{2}\\&=3x^{2}\end{alignedat}}}
f ( x ) = x n {\displaystyle f(x)=x^{n}}
d d x ( x n ) = lim h → 0 ( x + h ) n − x n h = lim h → 0 x n + n x n − 1 h + n ( n − 1 ) 2 ! x n − 2 h 2 + n ( n − 1 ) ( n − 2 ) 3 ! x n − 3 h 3 + . . . − x n h = lim h → 0 n x n − 1 h + n ( n − 1 ) 2 ! x n − 2 h 2 + n ( n − 1 ) ( n − 2 ) 3 ! x n − 3 h 3 + . . . h = lim h → 0 n x n − 1 + n ( n − 1 ) 2 ! x n − 2 h + n ( n − 1 ) ( n − 2 ) 3 ! x n − 3 h 2 + . . . = lim h → 0 n x n − 1 d d x ( x n ) = n x n − 1 {\displaystyle {\begin{alignedat}{4}{\frac {d}{dx}}(x^{n})\,\,&=\lim _{h\to 0}{\frac {(x+h)^{n}-x^{n}}{h}}\\&=\lim _{h\to 0}{\frac {x^{n}+nx^{n-1}h+{\frac {n(n-1)}{2!}}x^{n-2}h^{2}+{\frac {n(n-1)(n-2)}{3!}}x^{n-3}h^{3}+\,...-x^{n}}{h}}\\&=\lim _{h\to 0}{\frac {nx^{n-1}h+{\frac {n(n-1)}{2!}}x^{n-2}h^{2}+{\frac {n(n-1)(n-2)}{3!}}x^{n-3}h^{3}+\,...}{h}}\\&=\lim _{h\to 0}nx^{n-1}+{\frac {n(n-1)}{2!}}x^{n-2}h+{\frac {n(n-1)(n-2)}{3!}}x^{n-3}h^{2}+\,...\\&=\lim _{h\to 0}nx^{n-1}\\{\frac {d}{dx}}(x^{n})\,\,&=nx^{n-1}\end{alignedat}}}
f ( x ) = sin x {\displaystyle f(x)=\sin x}
lim h → 0 sin h h = 1 {\displaystyle \lim _{h\to 0}{\frac {\sin h}{h}}=1}
sin ( a + b ) = sin a cos b + cos a sin b {\displaystyle \sin(a+b)=\sin a\cos b+\cos a\sin b}
d d x ( sin x ) = lim h → 0 sin ( x + h ) − sin x h = lim h → 0 sin x cos h + cos x sin h − sin x h = lim h → 0 sin x cos h − 1 h + lim h → 0 cos x sin h h {\displaystyle {\begin{alignedat}{3}{\frac {d}{dx}}(\sin x)\,\,&=\lim _{h\to 0}{\frac {\sin(x+h)-\sin x}{h}}\\&=\lim _{h\to 0}{\frac {\sin x\cos h+\cos x\sin h-\sin x}{h}}\\&=\lim _{h\to 0}\sin x{\frac {\cos h-1}{h}}+\lim _{h\to 0}\cos x{\frac {\sin h}{h}}\\\end{alignedat}}}
{ sin 2 θ = 2 sin θ cos θ cos 2 θ = 2 cos 2 θ − 1 {\displaystyle {\begin{cases}\sin 2\theta =2\sin \theta \cos \theta \\\cos 2\theta =2\cos ^{2}\theta -1\\\end{cases}}}
{ cos 3 θ = 4 cos 3 θ − 3 cos θ sin 3 θ = 3 sin θ − 4 sin 3 θ {\displaystyle {\begin{cases}\cos 3\theta =4\cos ^{3}\theta -3\cos \theta \\\sin 3\theta =3\sin \theta -4\sin ^{3}\theta \\\end{cases}}}