if Y = cos x cube into sin square x find DY by DX
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By U.V rule,
d[cos(x)³×sin²(x)]/dx=
=cosx³.2sinx.cosx - sinx³.3x².sin²x
=cosx³.sin2x - 3x².sinx³.sin²x
This is the answer. First I didn't understand that it was cosx³ and not cos³x. But the above answer is now correct.(Using U.V rule and Chain rule of differentiation)
d[cos(x)³×sin²(x)]/dx=
=cosx³.2sinx.cosx - sinx³.3x².sin²x
=cosx³.sin2x - 3x².sinx³.sin²x
This is the answer. First I didn't understand that it was cosx³ and not cos³x. But the above answer is now correct.(Using U.V rule and Chain rule of differentiation)
kkn2:
i think missed differentiation of cos(x)^3 as -sin (x)^3 bro
no u missed sin^2( x)
Irs right there....but i missed the second term in uv rule.. just a sec
Now it's correct
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