Partial Derivative of f(x) with Sin(x^2)

[teng125]

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

does anybody knows how to solve this pls

 

[d_leet]

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]

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.

 

[teng125]

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]

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]

 

[teng125]

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]

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

 

[teng125]

okok...thanx

 

[quasar987]

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)

 

[teng125]

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]

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)

 

[Gagle The Terrible]

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

 

[quasar987]

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

 

[teng125]

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

 

[VietDao29]

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'.