# Partial derivative

f( x(1), ......, x(n) ) = sum (i=1) sin(x(i)^2) x(i)

does anybody knows how to solve this pls

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?

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.

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??

arildno
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There's nothing here to "solve".
You just have a function f, given by the prescription:
$$f(x_{1},\cdots{x}_{n})=\sum_{i=1}^{n}x_{i}\sin(x_{i}^{2})$$

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??

arildno
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It isn't an equation, it is a definition of the function f.

okok......thanx

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)

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

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)

This form of the expression makes it easier to "find" the derivatives, so I would write the equation this way.

quasar987
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Can you find the derivative of f(x) = xsin(x²) ?

i think can by using u'v + vu' formula rite??

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'.