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anonymous

  • 5 years ago

can anyone help me with this derivative? not too difficult, just a small problem with a constant... Need to find the derivative of 2x(cos 3x-1) with respect to x.

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  1. anonymous
    • 5 years ago
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    Begin by using the product rule: $$\frac{d}{dx}2x(\cos 3x -1) = (\cos 3x -1)\frac{d}{dx}2x + 2x\frac{d}{dx}(\cos 3x -1)$$ We can then get, applying the chain rule to the \(cos3x -1\) term: $$2(\cos 3x -1) + 2x(-3\sin3x) = 2\cos3x - 2 - 6x\sin 3x$$

  2. anonymous
    • 5 years ago
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    i keep getting -6x sin(3x-1) + 2cos (3x-1) and I know from the answer that those "1"'s are supposed to be 2's and I don't know how!

  3. anonymous
    • 5 years ago
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    Wait a minute... do you mean \(2x(\cos3x -1)\) or \(2x\;\cos(3x-1)\)? Those are two very different things...

  4. anonymous
    • 5 years ago
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    ok, i"m sorry let me start again so you know for definitely sure. I need to find the derivative of 2x(cos3x-1) with respect to x.

  5. anonymous
    • 5 years ago
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    Okay, then my answer should be correct.

  6. anonymous
    • 5 years ago
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    Let me study that. a minute, pls. These always throw me WAY off! thanks for the time, btw.

  7. anonymous
    • 5 years ago
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    That's right (like I doubted you!) lol. but I have such problems with the rules of derivatives when it comes to trig functions. Any suggestions on a good website that can act as a "cheat sheet" for me? I just can't keep them straight, if I know them at all!

  8. anonymous
    • 5 years ago
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    Well, you have to memorize the basic ones: cosine and sine. Fortunately, they're pretty easy: for since, it's cosine, and for cosine it's the negative sine. Then all the others are combinations of those through the rules: tangent is cosine/sine, for instance. That'll get you through the tough spots. But ultimately, by practicing enough, you'll be able to see them instantly. it's not going to be easy starting out, so don't expect it to be and you won't be disappointed. :-)

  9. anonymous
    • 5 years ago
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    How bout this one then? Need the derivative of sin 5x - 5x?

  10. anonymous
    • 5 years ago
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    Would it be d/dx of sin 5x - d/dx of 5x? That would be cos 5x - 5? I'm sure that's not right, but that's what i think!

  11. anonymous
    • 5 years ago
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    Close. You forgot to apply the chain rule to \(\sin 5x\).

  12. anonymous
    • 5 years ago
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    That's where I get stuck. I see that now. Do all the trig functions, the main ones really, act like that? I mean do they all pull that constant out in front? Like with the -3 in front of the sin x in the first equation I asked you about. And for the 5 in this one.

  13. anonymous
    • 5 years ago
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    It's not a matter of putting the constant in front. It's a matter of having nested functions. The trigonometric functions are functions of \(x\) alone: \(f(x) = \cos x\), \(f(x) = \sin x\), \(f(x) = \tan x\). As soon as what they're acting on is another function (i.e. something more complicated than just plain old \(x\)), you have to apply the chain rule. Suppose, fo instnace, that I have \(f(x) = \cos(3x^2)\). Here's how I do the derivative, in full detail: Let \(u = 3x^2\). Then we have: $$f(x) = \cos u$$ The derivative is therefore, by the chain rule: $$\frac{df}{dx} = \frac{df}{du}\frac{du}{dx}$$ Which is equal to: $$\frac{df}{dx} = -\sin u \times (6x)$$ But substituting back in, we get: $$\frac{df}{dx} = -6x\sin (3x^2)$$ I hope that clears things up a little.

  14. anonymous
    • 5 years ago
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    It does clear things up...a little. I am studying this and will take notes on your help. this calculus book needs to be burned or thrown out or something, because it just doesn' t speak my language! Thanks so much!

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