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trig equations, differentiation and chain rule

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Original post by keromedic
Is your '.' meant to represent multiplication? Use * for clarity in the future please :tongue: or learn latex.



. Is a perfectly acceptable symbol to represent multiplication
Original post by TenOfThem
. Is a perfectly acceptable symbol to represent multiplication


Fair enough :getmecoat:.
Original post by TenOfThem
No




I am stuck here.

First off; I started with cos(x)^1/2

=-sin(x)^1/2 (1/2x^-1/2)

I know that I multiply the -sin by the derivative but, I found myself confused by how to put it in practise
(edited 9 years ago)
Original post by apronedsamurai


I am stuck here.

First off; I started with cos(x)^1/2

=-sin(x)^1/2 (1/2x^-1/2)

I know that I multiply the -sin by the derivative but, I found myself confused by how to put it in practise


That is correct

−sin⁡x2x\dfrac{-\sin\sqrt{x}}{2\sqrt{x}}
Original post by TenOfThem
That is correct

−sin⁡x2x\dfrac{-\sin\sqrt{x}}{2\sqrt{x}}


And the answer is (-) sin over 2, because of the 1/2 coeffecient? (And overall negativity because of the presence of the - with the differentiation of the cos term).?
Original post by apronedsamurai
And the answer is (-) sin over 2, because of the 1/2 coeffecient? (And overall negativity because of the presence of the - with the differentiation of the cos term).?


Do you think the answer is

−sin⁡2\dfrac{-\sin}{2}

Because it isn't

I gave the answer in my last post

But yes, it is over 2 because of the half and negative because of differentiating cos(u)
Original post by TenOfThem
Do you think the answer is

−sin⁡2\dfrac{-\sin}{2}

Because it isn't

I gave the answer in my last post

But yes, it is over 2 because of the half and negative because of differentiating cos(u)


No, I said that as a shorthand to save me typing out the roots etc.
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Avatar for davros
davros
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Original post by apronedsamurai
This might sound a really stupid/random question but; when writing down sin and cos together; does sin ALWAYS come first?


If they are mutliplied together, the order doesn't matter, so

sinxcosx = cosxsinx

The same rule applies if we have more than 2 things multiplied together, so

sinx(2cosx) = 2sinxcosx = 2cosxsinx

I think this is where some of your confusion lies.

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