If f(x) = 1 x^2 + a^2 + x^2 + b^2 , then f'(x) is equal to - Vidyadip

MCQ

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

✓
${x \over {({a^2} - {b^2})}}\left[ {{1 \over {\sqrt {{x^2} + {a^2}} }} - {1 \over {\sqrt {{x^2} + {b^2}} }}} \right]$
B
${x \over {({a^2} + {b^2})}}\left[ {{1 \over {\sqrt {{x^2} + {a^2}} }} - {2 \over {\sqrt {{x^2} + {b^2}} }}} \right]$
C
${x \over {({a^2} - {b^2})}}\left[ {{1 \over {\sqrt {{x^2} + {a^2}} }} + {1 \over {\sqrt {{x^2} + {b^2}} }}} \right]$
D
$({a^2} + {b^2})\left[ {{1 \over {\sqrt {{x^2} + {a^2}} }} - {2 \over {\sqrt {{x^2} + {b^2}} }}} \right]$

✓

Answer

Correct option: A.

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

a
(a) $f(x) = \frac{1}{{\sqrt {{x^2} + {a^2}} + \sqrt {{x^2} + {b^2}} }}$

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

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

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

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

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