What is the differentiation of (sin3x-cos4x)(tan6x+sec8x)

expanded:

y = sin(4x)*sin(4x)*cos(3x)

need to use the product rule and chain rule

look at individual derivatives, use chain rule

d/dx sin(4x) = 4cos(4x)

d/dx cos(3x) = -3sin(3x)

put it together using product rule

dy/dx = 4cos(4x)*sin(4x)*cos(3x) + 4cos(4x)*sin(4x)*cos(3x) - 3sin(3x)*sin(4x)*sin(4x)

simplify

dy/dx = 8cos(4x)*sin(4x)*cos(3x) - 3sin(3x)*sin2(4x)

answer is A.

Hope it helps friend request me and I'll accept xx:)

3< x< 5

step-by-step explanation:

|x - 4| < 1

there are two solutions, one positive and one negative.   remember to flip the inequality when taking the negative solution

x-4 < 1       x-4 > -1

add 4 to each side

x-4+4 < 1+4       x-4+4 > -1+4

x< 5                 x> 3

3< x< 5