My university friend has challenged me to work out the above equation.
In fairness I did give him an equally difficult challenge on sport (as he is non-sporting).
It's just a fun thing that we are doing during lockdown.
Anyway, can somebody give me some help please as I haven't seen this type of equation before?
He says that he can give me another one instead, but I still need to know how to work them out first.
Many thanks.
What is the derivative of y=(3x^2+2) cos^2x ?
2020-05-29 12:08 am
回答 (3)
2020-05-29 9:48 am
y = ( 3x² + 2 ) cos² x
Suppose -- ( 3x² + 2 ) = f1(x)
cos² x = f2 (x)
then y = f1(x) * f 2 (x)
=> dy/dx = d f1(x)/dx * f2(x) + d f2(x)/dx * f1(x) ............. Product Rule
=> dy/dx = cos² x * ( 6 x ) + 2 cos x * (-sin x ) * ( 3x² + 2 )
=> dy/dx = cos² x * ( 6 x ) - ( 3x² + 2 ) (sin 2 x ) ..... since 2 sin x * cos x = sin 2x
=> dy/dx = ( 6 x ) cos² x - ( 3x² + 2 ) (sin 2 x ) ................. Answer
Suppose -- ( 3x² + 2 ) = f1(x)
cos² x = f2 (x)
then y = f1(x) * f 2 (x)
=> dy/dx = d f1(x)/dx * f2(x) + d f2(x)/dx * f1(x) ............. Product Rule
=> dy/dx = cos² x * ( 6 x ) + 2 cos x * (-sin x ) * ( 3x² + 2 )
=> dy/dx = cos² x * ( 6 x ) - ( 3x² + 2 ) (sin 2 x ) ..... since 2 sin x * cos x = sin 2x
=> dy/dx = ( 6 x ) cos² x - ( 3x² + 2 ) (sin 2 x ) ................. Answer
2020-05-29 3:09 am
y=(3x^2+2) (cosx)^2. Use product rule. Recall derivative of cos is -sin.
dy/dx = [ (cosx)^2 ][6x] + [3x^2+2][(2cosx)(-sinx)]
dy/dx = [ (cosx)^2 ][6x] + [3x^2+2][(2cosx)(-sinx)]
2020-05-29 12:25 am
Sorry. It wouldn't be fair to do your work for you. I'm sure your "Uni friend" would agree. Ask your friend to work it out in front of you. S/he shouldn't be asking questions without knowing the correct answer.
收錄日期: 2021-04-29 16:30:06
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