Answer:
[tex]F'(x) = \frac{x}{{1 +x^{2} } } + tan^{-1} (x)}[/tex]
Explanation:
Given - F(x) = xtan⁻¹x
To find - Differentiate using first principle
Formula used -
First Principal :
F'(x) = [tex]\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}[/tex]
and
[tex]tan^{-1}(A - B) = tan^{-1}(\frac{A - B}{1 + AB} )[/tex]
Proof -
Given that, F(x) = xtan⁻¹x
⇒F(x+h) = (x+h) tan⁻¹(x+h)
So,
[tex]F'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\\= \lim_{h \to 0} \frac{(x+h)tan^{-1} (x+h) - xtan^{-1} x}{h}\\= \lim_{h \to 0} \frac{xtan^{-1} (x+h) + htan^{-1} (x+h) - xtan^{-1} x}{h}\\= \lim_{h \to 0} \frac{x(tan^{-1} (x+h) - tan^{-1} x) + htan^{-1} (x+h)}{h}\\= \lim_{h \to 0} \frac{x}{h} (tan^{-1} (x+h) - tan^{-1} x) + tan^{-1} (x+h)}[/tex]
Now,
We know that,
[tex]tan^{-1}(A - B) = tan^{-1}(\frac{A - B}{1 + AB} )[/tex]
[tex]F'(x) = \lim_{h \to 0} \frac{x}{h} (tan^{-1} (x+h) - tan^{-1} x) + tan^{-1} (x+h)}\\= \lim_{h \to 0} \frac{x}{h} tan^{-1} (\frac{(x+h) - x}{1 + (x+h)x}) + tan^{-1} (x+h)}\\= \lim_{h \to 0} \frac{x}{h} tan^{-1} (\frac{h}{1 + (x+h)x}) + tan^{-1} (x+h)}\\\\= \lim_{h \to 0} \frac{x}{h\frac{1 + x(x+h)}{1 +x(x+h)} } tan^{-1} (\frac{h}{1 + (x+h)x}) + tan^{-1} (x+h)}\\= \lim_{h \to 0} \frac{x}{{1 +x(x+h)} }.\frac{1 +x(x+h)}{h} tan^{-1} (\frac{h}{1 + (x+h)x}) + tan^{-1} (x+h)}[/tex]
Now,
We know that,
[tex]\lim_{h \to 0} \frac{tan^{-1} x}{x} = 1[/tex]
∴ we get
[tex]F'(x) = \lim_{h \to 0} \frac{x}{{1 +x(x+h)} }.\frac{1 +x(x+h)}{h} tan^{-1} (\frac{h}{1 + (x+h)x}) + tan^{-1} (x+h)}\\ = \lim_{h \to 0} \frac{x}{{1 +x(x+h)} }.\lim_{h \to 0}\frac{1 +x(x+h)}{h} tan^{-1} (\frac{h}{1 + (x+h)x}) + \lim_{h \to 0}tan^{-1} (x+h)}\\= \frac{x}{{1 +x(x+0)} }.1 + tan^{-1} (x+0)}\\= \frac{x}{{1 +x(x)} } + tan^{-1} (x)}\\= \frac{x}{{1 +x^{2} } } + tan^{-1} (x)}[/tex]
So,
We get
[tex]F'(x) = \frac{x}{{1 +x^{2} } } + tan^{-1} (x)}[/tex]