Question

The quotient rule, (u/v)' is

(u'v-uv')/2
(u'v-uv')/v
(u'v+uv')/V2
(u'v-uv')/v2​

Answer 1

SOLUTION

TO CHOOSE THE CORRECT OPTION

$\displaystyle \sf{The \: quotient \: rule \: \bigg( \frac{u}{v} \bigg)' \: \: is \: }$

$\displaystyle \sf{ \: \: \frac{u'v - uv'}{2} \: \: \ }$

$\displaystyle \sf{ \: \: \frac{u'v - uv'}{v} \: \: \ }$

$\displaystyle \sf{ \: \: \frac{u'v + uv'}{ {v}^{2} } \: \: \ }$

$\displaystyle \sf{ \: \: \frac{u'v - uv'}{ {v}^{2} } \: \: \ }$

EVALUATION

The quotient rule in derivative gives :

If u(x) & v(x) are two functions of x with g(x) ≠ 0

$\displaystyle \sf{ \frac{d}{dx} \bigg( \frac{u}{v} \bigg) \: = \frac{u'v - uv'}{ {v}^{2} } \: \: \ }$

From above formula we can conclude that

$\displaystyle \sf{ \bigg( \frac{u}{v} \bigg)' \: = \frac{u'v - uv'}{ {v}^{2} } \: \: \ }$

FINAL ANSWER

Hence the correct option is

$\displaystyle \sf{ \bigg( \frac{u}{v} \bigg)' \: = \frac{u'v - uv'}{ {v}^{2} } \: \: \ }$

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