Prove that: $\frac{\text{d}}{\text{d}x}(sin x)= cos x$
Proof : According to the definition of derivative
$\frac{\text{d}y}{\text{d}x}=\lim_{\triangle x\rightarrow 0}\frac{f(x +\triangle x)-f(x)}{\triangle x}$
$\frac{\text{d}y}{\text{d}x}=\lim_{\triangle x\rightarrow 0}\frac{sin(x +\triangle x)-sin(x)}{\triangle x}$
= $\lim_{\triangle x\rightarrow 0}\frac{[sin(x)cos(\triangle x)+cos(x)sin(\triangle x)]-[sin(x)]}{\triangle x}$
=$\lim_{\triangle x\rightarrow 0}\frac{sin(x)cos(\triangle x)-sin(x) +cos(x)sin(\triangle x)}{\triangle x}$
= $\lim_{\triangle x\rightarrow 0}\frac{sin(x)cos(\triangle x)-sin(x) +cos(x)sin(\triangle x)}{\triangle x}$
= $\lim_{\triangle x\rightarrow 0}\frac{sin(x)[cos(\triangle x)-1] +cos(x)sin(\triangle x)}{\triangle x}$
=$\lim_{\triangle x\rightarrow 0}\frac{cos(x)sin(\triangle x) -sin(x)[cos(\triangle x)-1]} {\triangle x}$
=$ \lim_{\triangle x\rightarrow 0}\frac{cos(x)sin(\triangle x)}{\triangle x} -\frac{sin(x)[cos(\triangle x)-1]} {\triangle x}$
=$ \lim_{\triangle x\rightarrow 0}\frac{cos(x)sin(\triangle x)}{\triangle x} -\lim_{\triangle x \rightarrow 0}\frac{sin(x)[cos(\triangle x)-1]} {\triangle x}$
= $ cos(x)\lim_{\triangle x\rightarrow 0}\frac{sin(\triangle x)}{\triangle x} -sin(x)\lim_{\triangle x \rightarrow 0}\frac{[cos(\triangle x)-1]} {\triangle x}$
= cos(x).1 - 0
$\frac{\text{d}}{\text{d}x}(sin (x))= cos x$
Prove that: $\frac{\text{d}}{\text{d}x}(cos(x))= -sin(x) x$
Proof : According to the definition of derivative
$\frac{\text{d}y}{\text{d}x}=\lim_{\triangle x\rightarrow 0}\frac{f(x +\triangle x)-f(x)}{\triangle x}$
$\frac{\text{d}y}{\text{d}x}=\lim_{\triangle x\rightarrow 0}\frac{cos(x +\triangle x)-cos(x)}{\triangle x}$
= $\lim_{\triangle x\rightarrow 0}\frac{[cos(x)cos(\triangle x)-sin(x)sin(\triangle x)]-[cos(x)]}{\triangle x}$
= $\lim_{\triangle x\rightarrow 0}\frac{cos(x)cos(\triangle x)-cos(x) - sin(x)sin(\triangle x)}{\triangle x}$
=$\lim_{\triangle x\rightarrow 0}\frac{cos(x)[cos(\triangle x)-1] -sin(x)sin(\triangle x)}{\triangle x}$
=$ \lim_{\triangle x\rightarrow 0}\frac{cos(x)[cos(\triangle x)-1]}{\triangle x} -\frac{sin(x)[sin(\triangle x)} {\triangle x}$
=$ \lim_{\triangle x\rightarrow 0}\frac{cos(x)[cos(\triangle x)-1]}{\triangle x} -\lim_{\triangle x \rightarrow 0}\frac{sin(x)sin(\triangle x)} {\triangle x}$
= 0 - sin(x).1
= -sin(x)
$\frac{\text{d}}{\text{d}x}(cos (x))= -sin (x)$
Prove that :$ \frac{\text{d}}{\text{d}x}(tan(x))= sec^{2}x$
Proof : $\frac{\text{d}y}{\text{d}x}=\lim_{\triangle x\rightarrow 0}\frac{f(x +\triangle x)-f(x)}{\triangle x}$
$\frac{\text{d}y}{\text{d}x}=\lim_{\triangle x\rightarrow 0}\frac{tan(x +\triangle x)-tan(x)}{\triangle x}$
=$\lim_{\triangle x\rightarrow 0}\frac{\frac{tan(x)+tan(\triangle x)}{1-tan(x)tan(\triangle x)}-tan(x)}{\triangle x}$
=$\lim_{\triangle x\rightarrow 0}\frac{\frac{tan(x) +tan(\triangle x)-tan(x)(1-tan(x)tan(\triangle x)}{(1-tan(x)tan(\triangle x)}}{\triangle x}$
=$\lim_{\triangle x\rightarrow 0}\frac{\frac{tan(x) tan(\triangle x)-tan(x)+tan^{2}(x)tan(\triangle x)}{(1-tan(x)tan(\triangle x)}}{\triangle x}$
=$\lim_{\triangle x\rightarrow 0}\frac{{ tan(\triangle x)(1+tan^{2}(x)}{}}{(1-tan(x)tan(\triangle x))\triangle x}$
=$\lim_{\triangle x \rightarrow 0}\frac{tan(\triangle x)}{\triangle x}\lim_{\triangle x \rightarrow 0}\frac{sec^{2}(x)}{(1-tan(x)tan(\triangle x))}$
=$1.\frac{sec^{2}(x)}{(1-tan(x)tan(0)}$
=$1 \frac{sec^{2}(x)}{(1-tan(x).0}$
$\frac{\text{d}}{\text{d}x}(tan(x))= sec^{2}x$