what is derivative of sin^2 (3x-1)^2 ?
dy/dx (sin^2 (x))= 2 sin x cos x
dy/dx (3x-1)^2= 6(3x-1)
chain rule?
= 2 sin ((3x-1)^2) cos ((3x-1)^2) * 6(3x-1)
Please help. Thank you
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I'm not sure if this is correct. Can someone clarify thanks.
you are correct
Or, you could go on and say
sin (2(3x-1)^2) * 6(3x-1)
Or, you could go on and say
sin (2(3x-1)^2) * 6(3x-1)
I got 12 sin (3x-1)^2 cos (3x-1)^2 * (3x-1) as my final answer. Is this in the correct simplified form ?
it is ok, but you can use
2sinu cosu = sin(2u)
to make it what I had:
6(3x-1) sin(2(3x-1)^2)
If you want to retain your original argument of sin and cos, then leave it as you had it.
2sinu cosu = sin(2u)
to make it what I had:
6(3x-1) sin(2(3x-1)^2)
If you want to retain your original argument of sin and cos, then leave it as you had it.
I have taken that into consideration; thank you!
5 answers