Image Processing

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w2schools way2school
Posted January 6, 2019 at 10:04 pm | #
$\begin{align*}f(x)&=e^{\sqrt{cot~x}}\\f(x+h)&=e^{\sqrt{cot(x+h)}}\\\dfrac{d}{dx}f(x)&=lim_{h\to0}~\dfrac{f(x+h)-f(x)}{h}\\&=lim_{h\to0}~\dfrac{e^\sqrt{cot(x+h)}-e^\sqrt{cot~x}}{h}\\&=lim_{h\to0}~e^\sqrt{cot~x}~.\dfrac{e^{\sqrt{cot(x+h)}-\sqrt{cot~x}}-1}{h}\\&=lim_{h\to0}~e^\sqrt{cot~x}~.\Bigg(\dfrac{e^{\sqrt{cot(x+h)}-\sqrt{cot~x}}-1}{\sqrt{cot(x+h)}-\sqrt{cot~x}}\Bigg)\times\dfrac{\sqrt{cot(x+h)}-\sqrt{cot~x}}{h}\hspace{1cm}\Bigg[Since,~lim_{x\to0}~\dfrac{e^x-1}{x}=1\Bigg]\\&=lim_{h\to0}~e^\sqrt{cot~x}~.\dfrac{\sqrt{cot(x+h)}-\sqrt{cot~x}}{h}\\&=lim_{h\to0}~e^\sqrt{cot~x}~.\dfrac{\sqrt{cot(x+h)}-\sqrt{cot~x}}{h}\times\dfrac{\sqrt{cot(x+h)}+\sqrt{cot~x}}{\sqrt{cot(x+h)}+\sqrt{cot~x}}\hspace{1cm}[Rationalising~the~numerator]\\&=lim_{h\to0}~e^\sqrt{cot~x}~.\dfrac{{cot(x+h)}-{cot~x}}{h\times({\sqrt{cot(x+h)}}+\sqrt{cot~x})}\\&=lim_{h\to0}~e^\sqrt{cot~x}~.\dfrac{\dfrac{cot~(x+h)~cot~x+1}{cot~(x+h-x)}}{h\times({\sqrt{cot(x+h)}}+\sqrt{cot~x})}\hspace{1cm}\Bigg[cot~(A-B)=\dfrac{cot~A~cot~B+1}{cot~A-cot~B}\Bigg]\\&=lim_{h\to0}~e^\sqrt{cot~x}~.\dfrac{\dfrac{cot~(x+h)~cot~x+1}{cot~h}}{h\times({\sqrt{cot(x+h)}}+\sqrt{cot~x})}\\&=lim_{h\to0}~e^\sqrt{cot~x}~.\dfrac{{cot~(x+h)~cot~x+1}}{{cot~h}\times{h}\times({\sqrt{cot(x+h)}}+\sqrt{cot~x})}\\&=lim_{h\to0}~e^\sqrt{cot~x}~.\dfrac{{cot~(x+h)~cot~x+1}}{\dfrac{h}{tan~h}\times({\sqrt{cot(x+h)}}+\sqrt{cot~x})}\\&=lim_{h\to0}~e^\sqrt{cot~x}~.\dfrac{{cot~(x+h)~cot~x+1}}{\dfrac{h}{\dfrac{tan~h}{h}\times{h}}\times({\sqrt{cot(x+h)}}+\sqrt{cot~x})}\hspace{1cm}\Bigg[Since,~lim_{h\to0}~\dfrac{tan~h}{h}=1\Bigg]\\&=\dfrac{e^{\sqrt{cot}}\times(cot~x^2+1)}{2\sqrt{cot~x}}\hspace{1cm}\Bigg[Since,~(1+cot^2~x=cosec^2~x)\Bigg]\\&=\dfrac{e^{\sqrt{cot}}cosec^2~x}{2\sqrt{cot~x}}\end{align*}$
showing here image but not showing image on my webpage
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Image Processing

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