Let f x -log e u x - v x - u -2-4- v -2-2- u 2-2- v 2-1- then f -2 is equal toA- -1B- 0C- 1D- None of these

The correct option is B 0f(x) = log_e {u(x)/v(x)} = log_e u(x) log_ev(x) ∴ f'(x) = u'(x)/u(x) v'(x)/v(x) f'(2) = u'(2)/u(2) v'(2)/v(2) = 4/2 2/1 = 2 ...

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Byju's Answer

Standard XII

Mathematics

Addition Rule of Differentiation

Let f x =log ...

Question

$L e t f (x) = l o g_{e} {\frac{u (x)}{v (x)}}, u^{'} (2) = 4, v^{'} (2) = 2, u (2) = 2, v (2) = 1, t h e n f^{'} (2) i s e q u a l t o$

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- 1

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None of these

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Solution

$f (x) = l o g_{e} {\frac{u (x)}{v (x)}}$
$= l o g_{e} u (x) - l o g_{e} v (x)$
$∴ f^{'} (x) = \frac{u^{'} (x)}{u (x)} - \frac{v^{'} (x)}{v (x)}$
$f^{'} (2) = \frac{u^{'} (2)}{u (2)} - \frac{v^{'} (2)}{v (2)}$
$= \frac{4}{2} - \frac{2}{1}$
$= 2 - 2 = 0$

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Let f(x)=loge{u(x)v(x)},u′(2)=4,v′(2)=2,u(2)=2,v(2)=1,then f′(2) is equal to

f(x)=loge{u(x)v(x)} =loge u(x)−logev(x) ∴f′(x)=u′(x)u(x)−v′(x)v(x) f′(2)=u′(2)u(2)−v′(2)v(2) =42−21 =2−2=0

$L e t f (x) = l o g_{e} {\frac{u (x)}{v (x)}}, u^{'} (2) = 4, v^{'} (2) = 2, u (2) = 2, v (2) = 1, t h e n f^{'} (2) i s e q u a l t o$

$f (x) = l o g_{e} {\frac{u (x)}{v (x)}}$
$= l o g_{e} u (x) - l o g_{e} v (x)$
$∴ f^{'} (x) = \frac{u^{'} (x)}{u (x)} - \frac{v^{'} (x)}{v (x)}$
$f^{'} (2) = \frac{u^{'} (2)}{u (2)} - \frac{v^{'} (2)}{v (2)}$
$= \frac{4}{2} - \frac{2}{1}$
$= 2 - 2 = 0$