Differentiation with fractions and so many variables...

[gabrielwong1991]

Hi guys... I have this equation need to solve which differentiate this pi-A with respect to Pa... I dont know how to proceed after breaking the brackets.. It involves fraction and so many variables and seems to have chain rule and product rules in it...

Could somebody give me some hints on how to differentiate this?


The answer will be dPi-_A/dP_A =0 ==> P_A=8+2p_B-p_B²/14-5p_B ​....

 

[Ishuda]

Just so I can get it clear in my mind, I believe you have something like
R = [1 - (1-v)²]/P_A
where my R is your Pi-_A,
v = \(\displaystyle \frac{2 - 2 P_A + P_B}{4 - P_A - P_B}\),
and you want the derivative of R wrt P_A.

Your approach was to replace the v with its formula and multiply every thing out, then take the derivative. Might I suggest, you do, as you suggested, use the chain rule, etc. but, initially, just use the quotient rule and write the derivative as
dR/dP_A = [(1-v)² - 1 + 2 P_A (1-v) dv/dP_A] / P_A²
or possibly as
dR/dP_A = [-R + 2 (1-v) dv/dP_A] / P_A
and simplify that before making the substitution? If you want to have dR/dP_A equal to zero, that might even offer more simplifications.

 

[gabrielwong1991]

ok mate, thanks for your suggestion let me try it now!

It is actually

R = [1 - (1-v)²]*P_A ​instead of divided by Pa

Just so I can get it clear in my mind, I believe you have something like
R = [1 - (1-v)²]/P_A
where my R is your Pi-_A,
v = \(\displaystyle \frac{2 - 2 P_A + P_B}{4 - P_A - P_B}\),
and you want the derivative of R wrt P_A.

Your approach was to replace the v with its formula and multiply every thing out, then take the derivative. Might I suggest, you do, as you suggested, use the chain rule, etc. but, initially, just use the quotient rule and write the derivative as
dR/dP_A = [(1-v)² - 1 + 2 P_A (1-v) dv/dP_A] / P_A²
or possibly as
dR/dP_A = [-R + 2 (1-v) dv/dP_A] / P_A
and simplify that before making the substitution? If you want to have dR/dP_A equal to zero, that might even offer more simplifications.