Find the derivation of each of the following from the first principle:.
$\frac{5-x}{5+x}$
Let
$f(x)=\frac{5-x}{5+x}$
We need to find the derivative of f(x) i.e. f’(x)
We know that,
$\mathrm{f}^{\prime}(\mathrm{x})=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(\mathrm{x}+\mathrm{h})-\mathrm{f}(\mathrm{x})}{\mathrm{h}}$ …(i)
$f(x)=\frac{5-x}{5+x}$
$\mathrm{f}(\mathrm{x}+\mathrm{h})=\frac{5-(\mathrm{x}+\mathrm{h})}{5+(\mathrm{x}+\mathrm{h})}=\frac{5-\mathrm{x}-\mathrm{h}}{5+\mathrm{x}+\mathrm{h}}$
Putting values in (i), we get
$\mathrm{f}^{\prime}(\mathrm{x})=\lim _{\mathrm{h} \rightarrow 0} \frac{\frac{5-\mathrm{x}-\mathrm{h}}{5+\mathrm{x}+\mathrm{h}}-\frac{5-\mathrm{x}}{5+\mathrm{x}}}{\mathrm{h}}$
$=\lim _{\mathrm{h} \rightarrow 0} \frac{\frac{(5-\mathrm{x}-\mathrm{h})(5+\mathrm{x})-(5-\mathrm{x})(5+\mathrm{x}+\mathrm{h})}{(5+\mathrm{x}+\mathrm{h})(5+\mathrm{x})}}{\mathrm{h}}$
$=\lim _{h \rightarrow 0} \frac{25+5 x-5 x-x^{2}-5 h-x h-\left[25+5 x+5 h-5 x-x^{2}-x h\right]}{h(5+x+h)(5+x)}$
$=\lim _{h \rightarrow 0} \frac{25-x^{2}-5 h-x h-25-5 h+x^{2}+x h}{h(5+x+h)(5+x)}$
$=\lim _{h \rightarrow 0} \frac{-10 h}{h(5+x+h)(5+x)}$
$=\lim _{h \rightarrow 0} \frac{-10}{(5+x+h)(5+x)}$
Putting h = 0, we get
$=\frac{-10}{(5+x+0)(5+x)}$
$=\frac{-10}{(5+x)(5+x)}$
$=\frac{-10}{(5+x)^{2}}$
Hence,
$f^{\prime}(x)=\frac{-10}{(5+x)^{2}}$