$\frac{d}{d x} [s i n^{n} x . s i n . n x] =$

n sin^n-1 x.sin(n+1)x

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-n sin^n-1 x. sin(n+1)x

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nsin^n-1 x.sin(n-1)x

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nSinx

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Solution

The correct option is A
n sin^n-1 x.sin(n+1)x

$use the rule (u . v)^{1} = u^{1} v + u . v^{1}$

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(b) $\frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + x y = x$
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is equal to

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ddx[sinnx.sin.nx]=

The correct option is A n sinn-1 x.sin(n+1)x use the rule(u.v)1=u1v+u.v1

$\frac{d}{d x} [s i n^{n} x . s i n . n x] =$

The correct option is A
n sin^n-1 x.sin(n+1)x

$use the rule (u . v)^{1} = u^{1} v + u . v^{1}$