Given that fx-sin x-x- Then fx-

The correct option is C x cos x sin x/x^2Let g(x)=sin x and h(x)=1/x f(x)=g(x).h(x) f'(x)=g(x).h'(x)+g'(x).h(x) f'(x)=sin x. 1/x^2+cos x.1/x f'(x)=x cos x si ...

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Standard XI

Mathematics

Algebra of Derivatives

Given that fx...

Question

Given that $f (x) = \frac{s i n x}{x}$ . Then f'(x)=.

\frac{c o s x}{x}

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- \frac{s i n x}{x^{2}}

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\frac{x c o s x - s i n x}{x^{2}}

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\frac{x s i n x + c o s x}{x^{2}}

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Solution

The correct option is C $\frac{x c o s x - s i n x}{x^{2}}$
Let $g (x) = s i n x$ and $h (x) = \frac{1}{x}$
$f (x) = g (x) . h (x) f^{'} (x) = g (x) . h^{'} (x) + g^{'} (x) . h (x) f^{'} (x) = s i n x . \frac{- 1}{x^{2}} + c o s x . \frac{1}{x} f^{'} (x) = \frac{x c o s x - s i n x}{x^{2}}$

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Given that f(x)=sin xx. Then f'(x)=.

The correct option is C x cos x−sin xx2Let g(x)=sin x and h(x)=1x f(x)=g(x).h(x)f′(x)=g(x).h′(x)+g′(x).h(x)f′(x)=sin x.−1x2+cos x.1xf′(x)=x cos x−sin xx2

Given that $f (x) = \frac{s i n x}{x}$ . Then f'(x)=.

The correct option is C $\frac{x c o s x - s i n x}{x^{2}}$
Let $g (x) = s i n x$ and $h (x) = \frac{1}{x}$
$f (x) = g (x) . h (x) f^{'} (x) = g (x) . h^{'} (x) + g^{'} (x) . h (x) f^{'} (x) = s i n x . \frac{- 1}{x^{2}} + c o s x . \frac{1}{x} f^{'} (x) = \frac{x c o s x - s i n x}{x^{2}}$