find using first principles the derivative of cosx

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$$\\\displaystyle \lim_{h\rightarrow 0}\;\;\;\frac{f(x+h)-(x)}{h}\\\\ =\displaystyle \lim_{h\rightarrow 0}\;\;\;\frac{cos(x+h)-cos(x)}{h}\\\\ =\displaystyle \lim_{h\rightarrow 0}\;\;\;\frac{cosx*cosh-sinx*sinh-cos(x)}{h}\\\\ =\displaystyle \lim_{h\rightarrow 0}\;\;\;\frac{cosx(cosh-1)-sinx*sinh}{h}\\\\ =\displaystyle \lim_{h\rightarrow 0}\;\;\;\frac{cosx(cosh-1)}{h}-\frac{sinx*sinh}{h}\\\\ =\displaystyle \lim_{h\rightarrow 0}\;\;\;cosx*\frac{(cosh-1)}{h}-sinx*\frac{sinh}{h}\\\\ =cosx*0-sinx*1\\\\ =0-sin(x)\\\\ =-sin(x)$$

If you need these proved that also be done

$$\\\displaystyle \lim_{h\rightarrow 0}\;\;\;\frac{cosh-1}{h}=0\\\\ \displaystyle \lim_{h\rightarrow 0}\;\;\;\frac{sinh}{h}=1\\\\$$

find using first principles the derivative of cosx

$$\boxed{ e^{i\phi}= \cos{(\phi)}+i\sin{(\phi)} }$$

$$e^{i\phi}= \cos{(\phi)}+i\sin{(\phi)} \quad | \quad \frac {d}{dx}$$

$$\left( e^{i\phi} \right)' = ie^{i\phi} = \left( \cos{(\phi)}+i\sin{(\phi)} \right) ' = (\cos{(\phi)})'+i(\sin{(\phi)})'$$

$$ie^{i\phi} = (\cos{(\phi)})'+i(\sin{(\phi)})' \\ i\left( \cos{(\phi)}+i\sin{(\phi)} \right) = (\cos{(\phi)})'+i(\sin{(\phi)})' \\ i\cos{(\phi)}+\underbrace{i^2}_{i^2=-1}\sin{(\phi)} = (\cos{(\phi)})'+i(\sin{(\phi)})' \\ i\cos{(\phi)}-\sin{(\phi)} = (\cos{(\phi)})'+i(\sin{(\phi)})' \\ \textcolor[rgb]{1,0,0}{ -\sin{(\phi)} } + i\textcolor[rgb]{0,0,1}{\cos{(\phi)} }= \textcolor[rgb]{1,0,0}{ (\cos{(\phi)})'}+i\textcolor[rgb]{0,0,1} {(\sin{(\phi)})'}\\ \\ \textcolor[rgb]{1,0,0}{\boxed{ (\cos{(\phi)})'= -\sin{(\phi)} }} \\ \textcolor[rgb]{0,0,1}{\boxed{ (\sin{(\phi)})'= \cos{(\phi)} }} \\$$