Question

differentiate and simplify: y=[3(1-sinx)]/2cosx

Answer 1

take out the constant first...then quotient rule

Answer 2

ok thank you

Answer 3

im sorry, but can you explain further

Answer 4

y = ( 3 - 3sin(x) ) / 2cos(x)

y ' = (2cos(x) * (- 3cos(x) - (3 - 3sin(x)) * -2sin(x) ) / (2cos(x))^2

y ' = (-6cos^2(x) - (-6 + 6sin(x)) * sin(x) ) / (4cos^2(x))

y ' = (-6cos^2(x) - (-6sin(x) + 6sin^2(x)) ) / (4cos^2(x))

y ' = (-6cos^2(x) + 6sin(x) - 6sin^2(x)) ) / (4cos^2(x))

y ' = (-6cos^2(x) - 6sin^2(x) + 6sin(x) ) / (4cos^2(x))

y ' = (-6 ( cos^2(x) + sin^2(x) ) + 6sin(x) ) / (4cos^2(x))

y ' = (-6 ( 1 ) + 6sin(x) ) / (4cos^2(x))

y ' = (-6 + 6sin(x) ) / (4cos^2(x))

y ' = 2(-3 + 3sin(x) ) / (4cos^2(x))

y ' = (-3 + 3sin(x) ) / (2cos^2(x))
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Answer 5

thank you