If $f(x) = \frac{1}{{\sqrt {{x^2} + {a^2}} + \sqrt {{x^2} + {b^2}} }}$, then $f'(x)$ is equal to

Question

If $f(x) = \frac{1}{{\sqrt {{x^2} + {a^2}}  + \sqrt {{x^2} + {b^2}} }}$, then $f'(x)$ is equal to

Accepted Answer

$\frac{x}{{({a^2} - {b^2})}}\left[ {\frac{1}{{\sqrt {{x^2} + {a^2}} }} - \frac{1}{{\sqrt {{x^2} + {b^2}} }}} \right]$

Answer

$\frac{x}{{({a^2} + {b^2})}}\left[ {\frac{1}{{\sqrt {{x^2} + {a^2}} }} - \frac{2}{{\sqrt {{x^2} + {b^2}} }}} \right]$

Answer

$\frac{x}{{({a^2} - {b^2})}}\left[ {\frac{1}{{\sqrt {{x^2} + {a^2}} }} + \frac{1}{{\sqrt {{x^2} + {b^2}} }}} \right]$

Answer

$({a^2} + {b^2})\left[ {\frac{1}{{\sqrt {{x^2} + {a^2}} }} - \frac{2}{{\sqrt {{x^2} + {b^2}} }}} \right]$

differentiation