Partial derivative

  • Thread starter teng125
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  • #1
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f( x(1), ......, x(n) ) = sum (i=1) sin(x(i)^2) x(i)

does anybody knows how to solve this pls
 

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  • #2
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teng125 said:
f( x(1), ......, x(n) ) = sum (i=1) sin(x(i)^2) x(i)

does anybody knows how to solve this pls

What is this exactly? And what does this have to do with partial derivatives?
 
  • #3
quasar987
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Don't be intimidated by the sum sign. Write what the sum is explicitely and you'll find a more familiar form. Also, you can resolve the special case n=3 and renaming x(1)=1, x(2)=y and x(3)=z first before tackling the more abstract general case of n variables.
 
  • #4
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f( x(1), ......, x(n) ) = sin(x(1)^2) x(1) + sin(x(2)^2)x(2) + sin(x(2)^2)x(2) + sin(x(3)^2)x(3) + ...... +sin(x(n)^2)x(n)

should i write this eqn like this??
 
  • #5
arildno
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There's nothing here to "solve".
You just have a function f, given by the prescription:
[tex]f(x_{1},\cdots{x}_{n})=\sum_{i=1}^{n}x_{i}\sin(x_{i}^{2})[/tex]
 
  • #6
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so u mean i just have to write the eqn above and no need to do partial derivative as the question ask to do so??
 
  • #7
arildno
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It isn't an equation, it is a definition of the function f.
 
  • #8
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okok......thanx
 
  • #9
quasar987
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Now that you've expanded the sum, it should be easier to see what the derivative wrt x(1) is. (remember, the derivative of a sum is the sum of the derivative)
 
  • #10
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f( x(1), ......, x(n) ) = sin(x(1)^2) x(1) + sin(x(2)^2)x(2) + sin(x(2)^2)x(2) + sin(x(3)^2)x(3) + ...... +sin(x(n)^2)x(n)

so should i write this eqn like this??

for quasar987
 
  • #11
arildno
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Again, this isn't an equation, but a definition of a function
Secondly, try to formulate IN YOUR OWN WORDS what you are supposed to do with this function! (This, you have failed to do so far)
 
  • #12
This form of the expression makes it easier to "find" the derivatives, so I would write the equation this way.
 
  • #13
quasar987
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Can you find the derivative of f(x) = xsin(x²) ?
 
  • #14
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i think can by using u'v + vu' formula rite??
 
  • #15
VietDao29
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teng125 said:
i think can by using u'v + vu' formula rite??
No, no, no. Another wrong answer. It's uv'. Repeat after me: (uv)' = u'v + uv', (uv)' = u'v + uv'.
 

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