Question

Xtan-1x differentiate using first principle

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Answer 1

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

Step-by-step explanation:

Given - F(x) = xtan⁻¹x

To find - Differentiate using first principle

Formula used -

First Principal :

F'(x) =
`\lim_(h \to 0) (f(x+h) - f(x))/(h)`

and

`tan^(-1)(A - B) = tan^(-1)((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) (f(x+h) - f(x))/(h)\\= \lim_(h \to 0) ((x+h)tan^(-1) (x+h) - xtan^(-1) x)/(h)\\= \lim_(h \to 0) (xtan^(-1) (x+h) + htan^(-1) (x+h) - xtan^(-1) x)/(h)\\= \lim_(h \to 0) (x(tan^(-1) (x+h) - tan^(-1) x) + htan^(-1) (x+h))/(h)\\= \lim_(h \to 0) (x)/(h) (tan^(-1) (x+h) - tan^(-1) x) + tan^(-1) (x+h)}`

Now,

We know that,

`tan^(-1)(A - B) = tan^(-1)((A - B)/(1 + AB) )`

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

Now,

We know that,

`\lim_(h \to 0) (tan^(-1) x)/(x) = 1`

∴ we get

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