Answer:
1 / (1 + x²)
Step-by-step explanation:
Derivative of a function by first principle is:
[tex]f'(x)= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}[/tex]
Here, f(x) = tan⁻¹ x.
[tex]f'(x)= \lim_{h \to 0} \frac{tan^{-1}(x+h)-tan^{-1}x}{h}[/tex]
Use the difference of arctangents formula:
[tex]tan^{-1}a-tan^{-1}b=tan^{-1}(\frac{a-b}{1+ab})[/tex]
[tex]f'(x)= \lim_{h \to 0} \frac{tan^{-1}(\frac{x+h-x}{1+(x+h)x} )}{h}\\f'(x)= \lim_{h \to 0} \frac{tan^{-1}(\frac{h}{1+(x+h)x} )}{h}[/tex]
Next, we're going to use a trick by multiplying and dividing by 1+(x+h)x.
[tex]f'(x)= \lim_{h \to 0} \frac{tan^{-1}(\frac{h}{1+(x+h)x} )}{h}\frac{1+(x+h)x}{1+(x+h)x} \\f'(x)= \lim_{h \to 0} \frac{1}{1+(x+h)x} \frac{tan^{-1}(\frac{h}{1+(x+h)x} )}{\frac{h}{1+(x+h)x}}[/tex]
We can now evaluate the limit. We'll need to use the identity:
[tex]\lim_{x \to 0} \frac{tan^{-1}x}{x} =1[/tex]
This can be shown using squeeze theorem.
The result is:
[tex]f'(x)= \lim_{h \to 0} \frac{1}{1+(x+h)x}\\\frac{1}{1+x^2}[/tex]
