Why does xcosx simplify down to cosx-xsinx ?
wowee. i am going to assume you mean "why is the derivative of \[f(x)=x\cos(x)\] \[f'(x)=\cos(x)-x\sin(x)\]"
integration?
im assuming by simplify you mean integrates into
yes, that's what i meant satellite73
finding a derivative is by no means the same thing as "simplifying"
lol ... :) the derivative wins
could you please explain how it simplifies to that
or derives
it doesnt simplify to that ....
if anything it transforms into that, complicates into that, morphs into that .... by no means does it simplify into it
as the limit definition will attest to
a term that should be washed out of every idiot math teachers mouth ok done ranting. use the "product rule" \[(fg)'=f'g+g'f\] with \[f(x)=x,f'(x)=1,g(x)=\cos(x), f'(x)=-\sin(x)\] ad you will get it in one step
the product rule aint the "why" tho
a guy pulls a rabbit out of a hat, why? because he stuck his hand in the hat and pulled out a rabbit lol
i am making an assumption that the question is asking "what is the derivative of \[F(x)=x\cos(x)\] rather than what was actually asked. in other words i assume that the "why" meant "how do i get"
ok i got it thank you, thus using the product rule you can derive it to that.
btw i hope you know to write \[\cos(x)\] rather than the common but hideous \[\cos x\] it will save a lot of confusion when doing such problems
morphs . i like "it morphs into that"
my math teacher didnt give me the extra credit points; simply because I used IBP instead of a u-substitution. SHe said it was becasue she hasnt taught us that yet.
just call me morpheus lol
ibp??
oh right
integration by parts
so much for teachers encouraging initiative these days
heaven forbid you should know something right? meanwhile they all say "read ahead in the book, come prepared for class..."
but know only what i tell you. 0 is a natural number because i said so.
:)
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