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find the derivative

2007-10-14 17:00:39 · 5 answers · asked by Anonymous in Science & Mathematics Mathematics

5 answers

f'(u) = (u^2*1-u*2u)/u^4
f'(u) = (u^2-2u^2)/u^4
f'(u) = (-u^2)/u^4
f'(u) = -1/u^2

2007-10-14 17:09:16 · answer #1 · answered by ωĨŞΣ Ĝųγ 4 · 0 0

I'm assuming you mean f(u)=u/(u^2+10)

Let's rewrite f(u)=u*(u^2+10)^-1

Then we can use product rule:

f'(u) = 1*(u^2+10)^-1 - u * 2(u^2+10)^-2 * 2u

You can simplify that to a more presentable form by multiplying the (u^2+10)^-1 by (u^2+10)/(u^2+10)

Give it a shot. Good luck!


Also, check out: http://www.numberempire.com/derivatives.php

2007-10-14 17:15:47 · answer #2 · answered by Pete K 2 · 0 0

As you have written it, equation is:-
f ( u ) = ( u / u ² ) + 10
f ( u ) = 1 / u + 10
f ( u ) = u^(-1) + 10
f ` ( u ) = -1 u^(-2)
f ` ( u ) = - 1 / u ²
I only hope that you don`t mean
f ( u ) = u / (u ² + 10) which becomes a different type of question.

2007-10-14 20:16:00 · answer #3 · answered by Como 7 · 0 0

I'm gonna rewrite the problem how I'm used to it:

(x)dx/(x^2+10)
Let u = x^2 + 10
du = 2x dx
dx = du/2x
SO now its:

(x*du/2x)/(u)
1/2[du/u]
1/2[-u^-2]
-1/2[(x^2+10)^-2]
-1/(2(x^2+10)^2)

2007-10-14 17:12:18 · answer #4 · answered by Anonymous · 0 0

u/u^2 is really 1/u if u cancel out. this = u^-1
derivative is -u^-2 = -1/u^2

2007-10-14 17:10:32 · answer #5 · answered by Jamie G 2 · 0 0

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