Asked by ani
1)d/dx of cos4x/3-x^2(inculding quentient&chain rule)
2)sin^2x/1+cosx(same method)
2)sin^2x/1+cosx(same method)
Answers
Answered by
Reiny
#1
y = cos 4x/(3-x^2)
dy/dx = ( (3-x^2)(-4sin 4x) - cos 4x (-2x) )/(3-x^2)^2
try the second one, let me know what you got
y = cos 4x/(3-x^2)
dy/dx = ( (3-x^2)(-4sin 4x) - cos 4x (-2x) )/(3-x^2)^2
try the second one, let me know what you got
Answered by
Steve
you might be surprised at the answer to #2, unless you note that
sin^2/(1+cos) = (1-cos^2)/(1+cos) = 1-cos
no need for quotient rule or chain rule here.
sin^2/(1+cos) = (1-cos^2)/(1+cos) = 1-cos
no need for quotient rule or chain rule here.
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