Question

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

Answer 1

Hint- In order to solve this question, we have to simply differentiate $\dfrac{d}{{dx}}$ the $\sin x$ in order to get some variations to find out the ${n^{th}}$ derivative formula.

Complete Step-by-Step solution:
Assuming that,
 $y = \sin x - - - - - \left( 1 \right)$
Now, differentiating equation (1), we get
$\dfrac{{dy}}{{dx}} = \cos x - - - - - - \left( 2 \right)$
Again differentiating equation (2), we get
$\dfrac{{{d^2}y}}{{d{x^2}}} = - \sin x - - - - - - - \left( 3 \right)$
Again differentiating equation (3), we get
$\dfrac{{{d^{^3}}y}}{{d{x^3}}} = - \cos x - - - - - - - - \left( 4 \right)$
Again differentiating equation (4), we get
$\dfrac{{{d^4}y}}{{d{x^4}}} = \sin x - - - - - - - - (5)$
.
.
.
.
.
Therefore, differentiating equation (1) for ${n^{th}}$ , we get
$\dfrac{{{d^n}y}}{{d{x^n}}} = \sin \left( {\dfrac{{n\pi }}{2} + x} \right){\text{ + c }} - - - - - - \left( n \right)$
Now , let us check for the ${n^{th}}$ derivative by substituting the values of $n = 1,2,3......$
Put , $n = 1$
$\dfrac{{dy}}{{dx}} = \sin \left( {\dfrac{\pi }{2} + x} \right){\text{ + c = }}\cos x$
Positive Because $\sin $ lies in ${2^{nd}}$ quadrant
Put,$n = 2$
$\dfrac{{{d^2}y}}{{d{x^2}}} = \sin \left( {\pi + x} \right) = {\text{ - }}\sin x$ ,
negative sign because $\sin $ lies in ${3^{rd}}$ quadrant .
therefore our $\dfrac{{dy}}{{dx}},\dfrac{{{d^2}y}}{{d{x^2}}}$ ….. matches to equation $\left( 1 \right),\left( 2 \right),\left( 3 \right)$ , we get
Hence ,${n^{th}}$ derivative of $\sin x$ are
$\dfrac{{{d^n}y}}{{d{x^n}}} = \sin \left( {\dfrac{{n\pi }}{2} + x} \right){\text{ + c}}$

Note- whenever we face this type of question the key concept is that. We have to simply differentiate the $\sin x$ for max $4$ times to get some kind of variation and we have to check in which coordinate the $\sin x$ lies to get the sign convention.