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excluderho

excluderho

Answered question

2022-06-26

Why log 2 ( 2 sin ( π x ) ) d x log 3 ( 2 sin ( π x ) ) sin ( π x ) 3 π cos ( π x ) ?
Apparently I completely forgot the basics of calculus after 13 years of not studying it. Why won't:
d log 2 ( 2 sin ( π x ) ) d x = log 3 ( 2 sin ( π x ) ) sin ( π x ) 3 π cos ( π x )
UPDATE:
Sorry everybody, I messed up the question big time. Yes, originally it's an integration question why won't:
log 2 ( 2 sin ( π x ) ) d x = log 3 ( 2 sin ( π x ) ) sin ( π x ) 3 π cos ( π x )

Answer & Explanation

Hadley Cunningham

Hadley Cunningham

Beginner2022-06-27Added 20 answers

Using Chain rule:
d d x log 2 ( 2 sin ( π x ) ) = 2 log ( 2 sin ( π x ) ) × [ 2 × π × cos ( π x ) ] × 1 2 sin ( π x )
polivijuye

polivijuye

Beginner2022-06-28Added 16 answers

Let f ( x ) = 2 sin ( π x )
Then,
d log 2 ( f ( x ) ) d x = 2 log f ( x ) × ( log f ( x ) ) .
Here,
( log f ( x ) ) = f ( x ) f ( x ) = 2 cos ( π x ) × π 2 sin ( π x ) .
As a result, we have
d log 2 ( f ( x ) ) d x = 2 log ( 2 sin ( π x ) ) × 2 cos ( π x ) × π 2 sin ( π x ) .
If you feel any difficulty in my answer, just let me know.

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