Find derivative of..........?

FamousBrainy5

From: US
Posts: 16
Votes: 32
Can anyone explain me how to find derivative of 8 ^ (sin (8y) (or similar equation where sin x is as power of some number) thanks in advance

 

Answer No.1

MisterRoger1

From: US
Posts: 4
Votes: 8
D/dy[a^u] = a^u*ln(a)*du/dy. Here, u = sin(8y), so du/dy = 8cos(8y). a = 8, so d/dy[8^sin(8y))] = 8^(sin(8y))*ln(8)*8cos(8y), or 8ln(8)*8^(sin(8y))*cos(8y).

Answer No.2

LondonGuy

From: US
Posts: 25
Votes: 50
Let z = f(y) = 8^[sin(8y)] first take the natural logarithm log z = log [8^{sin(8y)}] = sin(8y) log 8 now you can differentiate with respect to y d log z / dy = log 8 d sin(8y) / dy now apply the chain rune on both sides: or, 1/z dz/dy = log 8 d sin (8y) / d(8y) * d(8y)/dy or, 1/z dz/dy = log 8 cos(8y) * 8 or, dz/dy = 8z log 8 cos(8y) now you can plug in the value of z from above dz/dy = 8 * 8^[sin(8y)] log 8 cos(8y) = 8^[sin(8y) + 1] log 8 cos(8y)

Answer No.3

FrenchSam

From: US
Posts: 9
Votes: 27
Let u = 8^[ sin(8y) ] log u = sin(8y) log 8 (1/u) ( du/dy ) = 8 cos (8y) log 8 du/dy = 8 u cos (8y) log 8 du/dy = ( 8 ) 8^[ sin(8y) ] cos (8y) log 8

Answer No.4

Anonymous

From: -
Posts: -
Votes: -
8^[sin(8y)] = e^[ln8 sin(8y)]it is good the anser is 8

Answer No.5

Anonymous

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Iam empressive

Answer No.6

Anonymous

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8^[sin(8y)] = e^[ln8 sin(8y)]it is good the anser is 8

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