Expert: Alon Mandes - 7/25/2009

Question
QUESTION: Hi Alon, I need to differentiate the following 8(5+x)/16-x2.
I have used the chain rule for the denominator and then leibniz quotient rule for the whole thing.I am finishing up with the denominator with a power of 4 whereas mathcad gives it as a power of 2.I feel i am over multiplying but cannot avoid it.Mathcad gives reult as 8/(16-x^2)+2(40+8x)x/(16-x^2)^2.Hope you can help

ANSWER: The rule for fractional derivative is :
[f/g]'=[f'g-g'f]/g^2 .
In our case : [ 8(5+x)/(16-x^2) ]' = 8 [ (5+x)/(16-x^2) ]'=
8[(5+x)'(16-x^2)-(16-x^2)'(5+x)]/(16-x^2)^2 =
8[(16-x^2)-(-2x)(5+x)]/(16-x^2)^2 =
8[16-x^2+10x+2x^2]/(16-x^2)^2 =
8[x^2+10x+16]/(16-x^2)^2 =
8[(x+8)(x+2)]/(16-x^2)^2 .

Please chick the Mathcad result again .

Alon.





---------- FOLLOW-UP ----------

QUESTION: You are good, now can you help me integrate (arctan x)/1+x^2?
I would be most pleased,

Thanks again, Dave.

ANSWER: 1st of all lets note that (Arctan[x])'=1/(1+x²). So, we have an expression from the form of
f(x)f'(x). This form can be achieved when differentiating f(x)² . Therefore :
∫Arctan[x]/(1+x²) dx = ½∫[2/(1+x²)]*Arctan[x] dx = ½Arctan²[x] .

Alon.

---------- FOLLOW-UP ----------

QUESTION: Hi Alon, thanks for your response but i am still a little lost.Can i use integration by parts using f(x)= arctan x and g'(x)=1/1+x^2??? Am i right in saying that the integral of g'(x) is ln(1+x^2)or simply 1/1+x^2.

using the by parts formula do we get

 -arctan*ln(1+x^2)+ arctan x*ln(1+x^2)

I am lost completely

Answer
Well, using integration by parts will give us the same result. You started correctly :
let's denote f(x) as Arctan[x] & g'(x) as 1/(1+x²), therefore : f'(x)=1/(1+x²) & g(x)=Arctan[x].
So,
∫ Arctan[x] * [1/(1+x²)] dx = Arctan²[x] - ∫ [1/(1+x²)] * Arctan[x] dx .
If we set ∫ Arctan[x] * [1/(1+x²)] dx = I then we have :
I=Arctan²[x]-I
2I=Arctan²[x]
I=½Arctan²[x] .

Alon.