Y -sin x - x -cos x- then dy - dx -A- sin 2 x-sin x-x cos x-cos 2 x-x-cos x2B- x cos x-cos 2 x-sin x-sin 2 x-x-cos x2C- none of theseD- x cos x-sin x-1-x-cos x2

The correct option is D xcosx sinx+1/(x+cosx)^2 Here u(x)=sin x, v(x)=x + cos x. dy/dx=(x+cosx)d(sin x)/dx sinxd(x+cos x)/dx/(x+cos x)^2 (Using quotient rule) ...

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Byju's Answer

Standard XI

Mathematics

Inequality

y =sin x / x ...

Question

$y = \frac{s i n x}{x + c o s x}, t h e n \frac{d y}{d x} = ?$

$\frac{x c o s x + c o s^{2} x - s i n x - s i n^{2} x}{(x + c o s x)^{2}}$

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$\frac{s i n^{2} x + s i n x - x c o s x - c o s^{2} x}{(x + c o s x)^{2}}$

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$\frac{x c o s x - s i n x + 1}{(x + c o s x)^{2}}$

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none of these

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Solution

The correct option is C
$\frac{x c o s x - s i n x + 1}{(x + c o s x)^{2}}$

Here u(x)=sin x, v(x)=x + cos x.
$\frac{d y}{d x} = \frac{(x + c o s x) \frac{d (s i n x)}{d x} - s i n x \frac{d (x + c o s x)}{d x}}{(x + c o s x)^{2}}$
(Using quotient rule)
= $\frac{(x + c o s x) c o s x - s i n x (1 - s i n x)}{(x + c o s x)^{2}} = \frac{x c o s x + c o s^{2} x - s i n x + s i n^{2} x}{(x + c o s x)^{2}}$
= $\frac{x c o s x - s i n x + s i n^{2} x + c o s^{2} x}{(x + c o s x)^{2}} = \frac{x c o s x - s i n x + 1}{(x + c o s x)^{2}}$

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y=sinxx+cosx, then dydx=?

The correct option is C xcosx−sinx+1(x+cosx)2 Here u(x)=sin x, v(x)=x + cos x. dydx=(x+cosx)d(sinx)dx−sinxd(x+cosx)dx(x+cosx)2 (Using quotient rule) =(x+cosx)cosx−sinx(1−sinx)(x+cosx)2=xcosx+cos2x−sinx+sin2x(x+cosx)2 =xcosx−sinx+sin2x+cos2x(x+cosx)2=xcosx−sinx+1(x+cosx)2

$y = \frac{s i n x}{x + c o s x}, t h e n \frac{d y}{d x} = ?$