Answer:
[tex]\dfrac{\text{d}}{\text{d}x} \tan^{-1}x=\dfrac{1}{1+x^2}[/tex]
Step-by-step explanation:
[tex]\boxed{\begin{minipage}{6.5 cm}\underline{Trigonometric Identity}\\\\$\tan^{-1}(A)-\tan^{-1}(B) \equiv \tan^{-1}\left(\dfrac{A-B}{1+AB}\right)$\\\end{minipage}}[/tex]
[tex]\boxed{\begin{minipage}{3 cm}$\displaystyle \lim_{h \to 0} \left[\dfrac{\tan^{-1} \theta}{\theta} \right]=1$\\\end{minipage}}[/tex]
[tex]\boxed{\begin{minipage}{5.6 cm}\underline{Differentiating from First Principles}\\\\$\text{f}\:'(x)=\displaystyle \lim_{h \to 0} \left[\dfrac{\text{f}(x+h)-\text{f}(x)}{(x+h)-x}\right]$\\\end{minipage}}[/tex]
Given function:
[tex]\text{f}(x)=\tan^{-1}x[/tex]
[tex]\implies \text{f}(x+h)=\tan^{-1}(x+h)[/tex]
Differentiating from first principles:
[tex]\begin{aligned}\text{f}\:'(x) & =\displaystyle \lim_{h \to 0} \left[\dfrac{\text{f}(x+h)-\text{f}(x)}{(x+h)-x}\right]\\\\& =\lim_{h \to 0} \left[\dfrac{\tan^{-1}(x+h)-\tan^{-1}x}{(x+h)-x}\right]\end{aligned}[/tex]
Using the trigonometric identity to rewrite the numerator:
[tex]\begin{aligned}& =\lim_{h \to 0} \left[\dfrac{\tan^{-1}\left(\dfrac{x+h-x}{1+x(x+h)}\right)}{(x+h)-x}\right]\\\\& =\lim_{h \to 0} \left[\dfrac{\tan^{-1}\left(\dfrac{h}{1+x^2+xh)}\right)}{h}\right]\end{aligned}[/tex]
[tex]\textsf{Multiply the denominator by }\dfrac{1+x^2+xh}{1+x^2+xh}:[/tex]
[tex]= \displaystyle \lim_{h \to 0} \left[\dfrac{\tan^{-1}\left(\dfrac{h}{1+x^2+xh)}\right)}{\dfrac{h(1+x^2+xh)}{(1+x^2+xh)}}\right][/tex]
Separate:
[tex]= \displaystyle \lim_{h \to 0} \left[\dfrac{\tan^{-1}\left(\dfrac{h}{1+x^2+xh)}\right)}{\dfrac{h}{(1+x^2+xh)}} \right] \cdot \displaystyle \lim_{h \to 0} \left[\dfrac{1}{1+x^2+xh}\right][/tex]
[tex]\textsf{Use }\displaystyle \lim_{h \to 0} \left[\dfrac{\tan^{-1} \theta}{\theta} \right]=1:[/tex]
[tex]= 1 \cdot \displaystyle \lim_{h \to 0} \left[\dfrac{1}{1+x^2+xh}\right][/tex]
As h gets close to zero:
[tex]= 1 \cdot \left[\dfrac{1}{1+x^2}\right][/tex]
Simplify:
[tex]=\dfrac{1}{1+x^2}[/tex]
