find derivative of log tan x using first principle.

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$$\begin{gathered} \mathrm{Given}, \\ \mathrm{f}\left(\mathrm{x}\right)=\log \left(\mathrm{tanx}\right) \\ \mathrm{f}\text{'}\left(\mathrm{x}\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{\mathrm{f}\left(\mathrm{x}+\mathrm{h}\right)-\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{h}} \\ =\underset{\mathrm{h}\to 0}{\lim }\frac{\log \left(\tan \left(\mathrm{x}+\mathrm{h}\right)\right)-\log \left(\mathrm{tanx}\right)}{\mathrm{h}} \\ =\underset{\mathrm{h}\to 0}{\lim }\frac{\log \left({\displaystyle \frac{\tan \left(\mathrm{x}+\mathrm{h}\right)}{\mathrm{tanx}}}\right)}{\mathrm{h}} \\ \mathrm{we} \mathrm{know} \mathrm{by} \mathrm{std}. \mathrm{form}, \underset{\mathrm{x}\to \mathrm{a}}{\lim }\frac{\ln \left(1+\mathrm{f}\left(\mathrm{x}\right)\right)}{\mathrm{f}\left(\mathrm{x}\right)}=1, \mathrm{where} \underset{\mathrm{x}\to \mathrm{a}}{\lim }\mathrm{f}\left(\mathrm{x}\right)=0 \\ \mathrm{so}, \\ =\underset{\mathrm{h}\to 0}{\lim }\frac{\log \left({\displaystyle 1+\frac{\tan \left(\mathrm{x}+\mathrm{h}\right)}{\mathrm{tanx}}-1}\right)}{\mathrm{h}} \\ \mathrm{let} \frac{\tan \left(\mathrm{x}+\mathrm{h}\right)}{\mathrm{tanx}}-1=\mathrm{f}\left(\mathrm{x}\right) \\ \mathrm{multiplying} \mathrm{and} \mathrm{dividing} \mathrm{by} \mathrm{f}\left(\mathrm{x}\right) \\ \underset{\mathrm{h}\to 0}{\lim }\frac{\log \left({\displaystyle 1+\frac{\tan \left(\mathrm{x}+\mathrm{h}\right)}{\mathrm{tanx}}-1}\right)}{\mathrm{h}\left(\frac{\tan \left(\mathrm{x}+\mathrm{h}\right)}{\mathrm{tanx}}-1\right)}\times \left(\frac{\tan \left(\mathrm{x}+\mathrm{h}\right)}{\mathrm{tanx}}-1\right) \\ \mathrm{since} , \underset{\mathrm{h}\to 0}{\lim }\left(\frac{\tan \left(\mathrm{x}+\mathrm{h}\right)}{\mathrm{tanx}}-1\right)=0, \mathrm{hence} \underset{\mathrm{h}\to 0}{\lim }\frac{\log \left({\displaystyle 1+\frac{\tan \left(\mathrm{x}+\mathrm{h}\right)}{\mathrm{tanx}}-1}\right)}{\left(\frac{\tan \left(\mathrm{x}+\mathrm{h}\right)}{\mathrm{tanx}}-1\right)}=1 \\ \underset{\mathrm{h}\to 0}{\lim }\frac{\left(\frac{\tan \left(\mathrm{x}+\mathrm{h}\right)}{\mathrm{tanx}}-1\right)}{\mathrm{h}} \\ \frac{1}{\mathrm{tanx}}\underset{\mathrm{h}\to 0}{\lim }\left(\frac{\tan \left(\mathrm{x}+\mathrm{h}\right)-\mathrm{tanx}}{\mathrm{h}}\right) \\ \frac{1}{\mathrm{tanx}}\underset{\mathrm{h}\to 0}{\lim }\frac{1}{\mathrm{h}}\left[\frac{\sin \left(\mathrm{x}+\mathrm{h}\right)}{\cos \left(\mathrm{x}+\mathrm{h}\right)}-\frac{\mathrm{sinx}}{\mathrm{cosx}}\right] \\ \frac{1}{\mathrm{tanx}}\underset{\mathrm{h}\to 0}{\lim }\frac{1}{\mathrm{h}}\left[\frac{\sin \left(\mathrm{x}+\mathrm{h}\right)\mathrm{cosx}-\mathrm{sinx}.\cos \left(\mathrm{x}+\mathrm{h}\right)}{\cos \left(\mathrm{x}+\mathrm{h}\right).\mathrm{cosx}}\right] \\ =\mathrm{using} \mathrm{formula} \mathrm{for} \sin \left(\mathrm{A}+\mathrm{B}\right) \\ =\frac{1}{\mathrm{tanx}}\underset{\mathrm{h}\to 0}{\lim }\frac{1}{\mathrm{h}}\frac{\sin \left(\mathrm{x}+\mathrm{h}-\mathrm{x}\right)}{\mathrm{hcos}\left(\mathrm{x}+\mathrm{h}\right)\mathrm{cosx}} \\ =\frac{1}{\mathrm{tanx}}\underset{\mathrm{h}\to 0}{\lim }\frac{\sinh }{\mathrm{h}}.\frac{1}{\cos \left(\mathrm{x}+\mathrm{h}\right)\mathrm{cosx}} \\ \mathrm{we} \mathrm{know}, \underset{\mathrm{h}\to 0}{\lim }\frac{\sinh }{\mathrm{h}}=1 \\ \mathrm{hence}, \\ \frac{1}{\mathrm{tanx}}\underset{\mathrm{h}\to 0}{\lim }\frac{1}{\cos \left(\mathrm{x}+\mathrm{h}\right)\mathrm{cosx}} \\ \frac{1}{\mathrm{tanx}.{\cos }^2\mathrm{x}}=\frac{{\sec }^2\mathrm{x}}{\mathrm{tanx}} \\ \mathrm{hence} , \\ \frac{\mathbf{d}}{\mathbf{dx}}\mathbf{log}\left(\mathbf{tanx}\right)\mathbf{=}\frac{{\mathbf{sec}}^{\mathbf{2}}\mathbf{x}}{\mathbf{tanx}} \end{gathered}$$

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