What-is-the-nth-derivative-of-sinx-in-terms-of-the-sine-function-

Question Number 59838 by necx1 last updated on 15/May/19

W h a t i s t h e n t h d e r i v a t i v e o f s i n x i n

t e r m s o f t h e s i n e f u n c t i o n ?

Commented by maxmathsup by imad last updated on 15/May/19

i f f (x) = s i n x, f^{(n)} (x) = s i n (x + \frac{n π}{2}) w i t h n ⩾ 1 a n d f^{(0)} = f

l e t p r o v e t h i s b y r e c u r r e n c e w e h a v e f^{(1)} (x) = c o s (x) = s i n (x + \frac{π}{2})

l e t s u p p o s e t h e e q u a l i t y t r u e a t t e r m n \Rightarrow f^{(n + 1)} (x) = {(f^{(n)} (x))}^{^{'}}

= {(s i n (x + \frac{n π}{2}))}^{^{'}} = c o s (x + \frac{n π}{2}) = s i n (x + \frac{n π}{2} + \frac{π}{2}) = s i n (x + \frac{(n + 1) π}{2})

t h e r e s u l t i s p r o v e d .

Answered by tanmay last updated on 15/May/19

y = s i n x

\frac{d y}{d x} = y_{1} = c o s x = s i n (\frac{π}{2} + x)

y_{2} = - s i n x = s i n (2 \times \frac{π}{2} + x)

y_{3} = - c o s x = s i n (3 \times \frac{π}{2} + x)

y_{4} = s i n x = s i n (4 \times \frac{π}{2} + x)

y_{n} = s i n (n \times \frac{π}{2} + x)

h e r e n = 9

y_{9} = s i n (9 \times \frac{π}{2} + x)

Commented by necx1 last updated on 15/May/19

o h \dots . . I g e t i t n o w . T h a n k s

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