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# nee help with this differentiation question watch

1. Its asking me to differentiate the function f(x) = 1/(x^2 + a^2)

am I right in thinking that I need to change it to partial fractions form?
2. (Original post by fuzzybear)
Its asking me to differentiate the function f(x) = 1/(x^2 + a^2)

am I right in thinking that I need to change it to partial fractions form?
Just use the quotient rule. Is a a constant?
3. (Original post by keromedic)
Just use the quotient rule. Is a a constant?
yeah 'a' is a constant
4. (Original post by keromedic)
Just use the quotient rule. Is a a constant?
I don't understand though, I thought the quotient rule can only be used when you have a function on the top and bottom of a fraction

in my question there, the top part of the fraction is just '1', there are no 'x's' on the top
5. Here you go.

http://i.imgur.com/pCRGCoi.jpg
6. f(x) = 1/x^2 + 4^2 (where a = 4)

f(x) = (x^2 + 4^2) ^ -1

dy/dx = -2x(x^2 + 4^2)^-2

Move the denominator to the numerator and use the chain rule.

7. (Original post by fuzzybear)
I don't understand though, I thought the quotient rule can only be used when you have a function on the top and bottom of a fraction

in my question there, the top part of the fraction is just '1', there are no 'x's' on the top
You do have a function on top, y=1 is still a function. You don't need the quotient rule though (but it would work), the chain rule is enough.
8. (Original post by fuzzybear)
I don't understand though, I thought the quotient rule can only be used when you have a function on the top and bottom of a fraction

in my question there, the top part of the fraction is just '1', there are no 'x's' on the top
The '1' can be considered a function, it's just a constant function, so d(1)/dx = 0. You can also do what the HazimQ did (chain rule).
9. (Original post by HazimQ)
f(x) = 1/x^2 + 4^2 (where a = 4)

f(x) = (x^2 + 4^2) ^ -1

dy/dx = -2x(x^2 + 4^2)^-2

Move the denominator to the numerator and use the chain rule.

Your first line is very ambiguous. Use brackets or Latex
10. (Original post by fuzzybear)
I don't understand though, I thought the quotient rule can only be used when you have a function on the top and bottom of a fraction

in my question there, the top part of the fraction is just '1', there are no 'x's' on the top
A function can be constant, i.e independent of x. The quotient rule still applies
11. I applied the quotient rule and got:

-2x / (x^2 + a^2)^2

-2x / x^4 + 2*(a^2)*(x^2) + a^4

is that correct?
12. (Original post by fuzzybear)
I applied the quotient rule and got:

-2x / (x^2 + a^2)^2

-2x / x^4 + 2*(a^2)*(x^2) + a^4

is that correct?
Indeedy. Expanding the bracket makes it ugly, but either way its right.
13. (Original post by Phichi)
Indeedy. Expanding the bracket makes it ugly, but either way its right.
sweet, cheers mate

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