[Blindmath] Dirivitives

[Joseph Lee]

Hi Jacques,
As someone pointed out, dy sqrt(t) is not right. For derivatives involving
products, quotients and composite function (chain rule), I'd start with
writing derivatives of each function first before invoking the formula like:
p(t) = sqrt(t)/(t-1)
T(high) = sqrt(-1^2)/2
t(low) = 1
Then:
dy(f/g) = (low*d high)-(high*d low)/(low(^2)
= ((t-1)*(sqrt t^-1/2/2))-((sqrt t)*1)/(t-1)^2
Leave denominator i.e. (t-1)^2 alone (which becomes t^2-2t+1 if you want an
exact answer)), then simplify the numerator one piece at a time - that is,
solve low*d high first, then solve high*d low, simplify both expressions and
combine i.e. subtract high*d low from low*d high.
BTW, start with and easier expression - in our case, high * d low, set this
aside for now and solve low*d high. For chain rule (which you may learn
soon), start with the inner function first, as the outer function will
depend on it (dy chain rule = dg(f(df))).
Cheers,
Joseph

-----Original Message-----
From: Blindmath [mailto:blindmath-bounces at nfbnet.org] On Behalf Of Jacques
Chappell via Blindmath
Sent: Saturday, October 18, 2014 11:20 AM
To: 'Blind Math list for those interested in mathematics'
Subject: [Blindmath] Dirivitives

I have a problem I have pasted into this email below, using the quotient
rule we are to find the dirivitive, but I am stuck. So I have the work I
have done so far and if I am correct or not correct in what I have done
please let me know. Also below that is the correct answer, but  I haven't
reached the answer yet. As I have said I am stuck. Thanks.

21) p(t)=sqrt t/t-1

(t-1)(t^1/2)-(sqrt t)(1)/(t-1)^2

T^3/2-t^1/2-sqrt t/(t-1)^2

T^3/2-t^1/2-t^1/2/(t-1)^2

T^3/2-2t^1/2/(t-1)^2

 

P'(t)=[-sqrt t/2-1/(2*sqrt t)]/(t-1)^2

 

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