Question

xtan-1x differentiate using first principle

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Answer 1

Answer:

$F'(x) = \frac{x}{{1 +x^{2} } } + tan^{-1} (x)}$

Explanation:

Given - F(x) = xtan⁻¹x

To find - Differentiate using first principle

Formula used -

First Principal :

F'(x) = $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

and

$tan^{-1}(A - B) = tan^{-1}(\frac{A - B}{1 + AB} )$

Proof -

Given that, F(x) = xtan⁻¹x

⇒F(x+h) = (x+h) tan⁻¹(x+h)

So,

$F'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\\= \lim_{h \to 0} \frac{(x+h)tan^{-1} (x+h) - xtan^{-1} x}{h}\\= \lim_{h \to 0} \frac{xtan^{-1} (x+h) + htan^{-1} (x+h) - xtan^{-1} x}{h}\\= \lim_{h \to 0} \frac{x(tan^{-1} (x+h) - tan^{-1} x) + htan^{-1} (x+h)}{h}\\= \lim_{h \to 0} \frac{x}{h} (tan^{-1} (x+h) - tan^{-1} x) + tan^{-1} (x+h)}$

Now,

We know that,

$tan^{-1}(A - B) = tan^{-1}(\frac{A - B}{1 + AB} )$

$F'(x) = \lim_{h \to 0} \frac{x}{h} (tan^{-1} (x+h) - tan^{-1} x) + tan^{-1} (x+h)}\\= \lim_{h \to 0} \frac{x}{h} tan^{-1} (\frac{(x+h) - x}{1 + (x+h)x}) + tan^{-1} (x+h)}\\= \lim_{h \to 0} \frac{x}{h} tan^{-1} (\frac{h}{1 + (x+h)x}) + tan^{-1} (x+h)}\\\\= \lim_{h \to 0} \frac{x}{h\frac{1 + x(x+h)}{1 +x(x+h)} } tan^{-1} (\frac{h}{1 + (x+h)x}) + tan^{-1} (x+h)}\\= \lim_{h \to 0} \frac{x}{{1 +x(x+h)} }.\frac{1 +x(x+h)}{h} tan^{-1} (\frac{h}{1 + (x+h)x}) + tan^{-1} (x+h)}$

Now,

We know that,

$\lim_{h \to 0} \frac{tan^{-1} x}{x} = 1$

∴ we get

$F'(x) = \lim_{h \to 0} \frac{x}{{1 +x(x+h)} }.\frac{1 +x(x+h)}{h} tan^{-1} (\frac{h}{1 + (x+h)x}) + tan^{-1} (x+h)}\\ = \lim_{h \to 0} \frac{x}{{1 +x(x+h)} }.\lim_{h \to 0}\frac{1 +x(x+h)}{h} tan^{-1} (\frac{h}{1 + (x+h)x}) + \lim_{h \to 0}tan^{-1} (x+h)}\\= \frac{x}{{1 +x(x+0)} }.1 + tan^{-1} (x+0)}\\= \frac{x}{{1 +x(x)} } + tan^{-1} (x)}\\= \frac{x}{{1 +x^{2} } } + tan^{-1} (x)}$

So,

We get

$F'(x) = \frac{x}{{1 +x^{2} } } + tan^{-1} (x)}$