Question

differentiate w.r.t x

Answer 1

this is the answer.......

Answer 2

Assuming log has base 'e'

Final Answer :
$\frac{1}{ \sqrt{ {x}^{2} + {a}^{2} } }$

Steps:
1) Let
$y = log( x+ \sqrt{ {x}^{2} + {a}^{2} } )$

Then, we know that
$\frac{dy}{dx} = \frac{1}{x + \sqrt{ {x}^{2} + {a}^{2} } } \times \frac{d(x + \sqrt{ {x}^{2} + {a}^{2} }) }{dx} \\ \\ = > \frac{1 + \frac{2x}{2 \sqrt{ {x}^{2} + {a}^{2} } } }{x + \sqrt{ {x}^{2} + {a}^{2} } } \\ = > \frac{1}{ \sqrt{ {x}^{2} + {a}^{2} } }$