Suppose x &#x2208;<!-- ∈ --> [ 0 , &#x03C0;<!-- π --> ] , let f

Yahir Tucker

Yahir Tucker

Answered question

2022-06-18

Suppose x [ 0 , π ], let f n ( x ) = cos n ( x ) sin ( x ), then the derivative is equal to:
cos n + 1 ( x ) n sin 2 ( x ) cos n 1 ( x ) = cos n 1 ( x ) [ ( n + 1 ) cos 2 ( x ) n ]
Could someone explain to me, how is the right part of the equation found?

Answer & Explanation

podesect

podesect

Beginner2022-06-19Added 20 answers

Note that
cos n + 1 x n sin 2 x cos n 1 x
= cos n 1 x ( cos 2 x n sin 2 x )
= cos n 1 x [ cos 2 x n ( 1 cos 2 x ) ]
= cos n 1 x ( cos 2 x n + n cos 2 x )
= cos n 1 x [ ( n + 1 ) cos 2 x n ]
Santino Bautista

Santino Bautista

Beginner2022-06-20Added 5 answers

cos n + 1 ( x ) n sin 2 ( x ) cos n 1 ( x ) = cos n + 1 ( x ) n ( 1 cos 2 ( x ) ) cos n 1 ( x ) = cos n + 1 ( x ) n cos n 1 ( x ) + n cos n + 1 ( x ) =

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