Question

ahoy!!

answer this!

differentiate w.r.t x :

=>
$log \: (x + \sqrt{ {x}^{2} + {a}^{2} } )$
thanks !!

Answer 1

hope it is correct................

Answer 2

$= \ \textgreater \ \frac{d}{dx} log( x + \sqrt{x^2 + a^2} )$

We know that log(x) = 1/x.

$= \ \textgreater \ \frac{1}{ \sqrt{x^2 + a^2} + x } * \frac{d}{dx} [ \sqrt{x^2 + a^2} + x]$

$= \ \textgreater \ \frac{ \frac{d}{dx} \sqrt{(x^2 + a^2)} + \frac{d}{dx} [x] }{ \sqrt{x^2 + a^2} + x}$

$= \ \textgreater \ \frac{ \frac{d}{dx} (x^2 + a^2)^ \frac{1}{2} + \frac{d}{dx} x }{ \sqrt{x^2 + a^2} + x }$

Note: We know that d/dx(x^1/2) = 1/2x^(1/2 - 1)

$= \ \textgreater \ \frac{ \frac{1}{2} (x^2 + a^2)^{ \frac{1}{2} - 1 } * \frac{d}{dx} [x^2 + a^2] + 1 }{ \sqrt{x^2 + a^2} + x }$  ------ (1)

$= \ \textgreater \ \frac{ \frac{1}{2} (x^2 + a^2)^{ \frac{1}{2} - 1 } * 2x }{ \sqrt{x^2 + a^2} + x }$

$= \ \textgreater \ \frac{ (x^2 + a^2)^{ \frac{1}{2} - 1 } x }{ \sqrt{x^2+a^2} + x }$

$= \ \textgreater \ \frac{x(a^2 + x^2)^ -\frac{1}{2} }{ \sqrt{x^2 + a^2} + x }$

$= \ \textgreater \ \frac{ \frac{x}{ \sqrt{x^2 + a^2} } }{ \sqrt{x^2 + a^2} + x } ----- (2)$

Substitute (2) in (1), we get

$= \ \textgreater \ \frac{ \frac{x}{ \sqrt{x^2 + a^2}} + 1 }{ \sqrt{x^2 + a^2} + x }$


$= \ \textgreater \ \frac{x+ \sqrt{a^2 + x^2} }{(x + \sqrt{a^2 + x^2) \sqrt{a^2 + x^2} } }$

$= \ \textgreater \ \frac{1}{ \sqrt{a^2 + x^2} }$



Hope this helps!