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The $n^{th}$ derivative of $\sin x$ is equal to-

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\sin (2 \frac{nπ}{2} + x) + c

\cos (\frac{nπ}{2} + x) + c

\sin (\frac{nπ}{2} + 2 x) + c

\sin (\frac{nπ}{2} + x) + c

answer is D.

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Detailed Solution

Let us assume that,

y = \sin x \dots \dots \dots (1)

Differentiating the equation (1), we get

\frac{dy}{dx} = \cos x \dots \dots \dots (2)

Equation (2) differentiating again, we get

\frac{d^{2} y}{d x^{2}} = - \sin x \dots \dots \dots (3)

Equation (3) differentiating again, we get

\frac{d^{3} y}{d x^{3}} = - \cos x \dots \dots \dots (4)

Equation (4) differentiating again, we get

\frac{d^{4} y}{d x^{4}} = \sin x \dots \dots \dots (5)

Therefore, differentiating for

n^{th}

we get

\frac{d^{n} y}{d x^{n}} = si n (\frac{nπ}{2} + x) + c \dots \dots \dots (n)

Now, let us check for the

n^{th}

derivative by substituting the values of

n = 1, 2, 3 \dots \dots

Put, n=1

\frac{dy}{dx} = \sin (\frac{π}{2} + x) + c = \cos x

Positive Because sin lies in

2^{nd}

quadrant
Put, n=2

\frac{d^{2} y}{d x^{2}} = \sin (π + x) = - \sin x

,
negative sign because sin lies in

3^{rd}

quadrant.
Therefore our

\frac{dy}{dx}, \frac{d^{2} y}{d x^{2}}

matches to equation (1), (2), (3), we get
Hence,

n^{th}

derivative of

\sin x

are

\frac{d^{n} y}{d x^{n}} = \sin (\frac{nπ}{2} + x) + c

Hence, correct option is 4.

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