posted
I've been learning Calculus for no particular reason, save pleasure. The pleasure aspect is, I must admit, quite minor at this point, but I'm hoping for great dividends when I finally finish, in a month or two.
Anyways, I've been working on this one problem for the last 30 minutes :/.
Find the derivative of: Sin^2(3x - 2)
I thought to myself immediately "power chain rule" and rushed to apply it. I tried breaking it up as follows:
Let t = 3x - 2
f(x) = Sin(t) * Sin(t) f'(x) = 3Cos(3x-2) * 3Cos(3x-2) Then, using the mulptication of derivatives rule,
a*b = a'b b'a
So...
3Cos(t)*(-)3Sin... I'll stop here, because this appears to already be far away from the answer given in the book.
I also tried turning it into a square problem:
Sin(t) = w
f(x) = w^2 f'(x) = 2w
and break it down from there, but that also doesn't seem to work.
I'm really frustrated, and would love some elucidation. Thanks .
Posts: 3060 | Registered: Nov 2003
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posted
As you note, this can be transformed to u^2.
The derivative of u^2 is, of course, 2u du/dx.
So we now have 2 sin(3x - 2) d(sin(3x - 2))/dx
the derivative of sin v is cos v dv/dx, so the derivative of sin(3x - 2) is 3 cos(3x - 2)
So the derivative of the whole thing is: 6 sin(3x - 2) cos(3x - 2)
This may look different due to application of some trigonometric identities. It is, however, definitely correct. I have verified it with my TI-92, just to be on the safe side (since once one's far enough past derivative they start to go all foggy -- most later derivatives can be handled just by throwing the fundamental theorem of calculus at them).
Posts: 15770 | Registered: Dec 2001
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