[Blindmath] Derivatives Help

[Jacques Chappell]

Wow thank you so much for such a detailed explanation of what to do. I
greatly appreciate it. That was awesome. Thank you.

-----Original Message-----
From: Blindmath [mailto:blindmath-bounces at nfbnet.org] On Behalf Of Sina
Bahram via Blindmath
Sent: Saturday, October 18, 2014 3:15 PM
To: 'Blind Math list for those interested in mathematics'; 'Sarah Jevnikar'
Subject: Re: [Blindmath] Derivatives Help

I was trying to write this without really doing it, but I'll settle for
trying to explain it well instead.

This is a simple application of the product rule. The original f(x) and g(x)
do not matter what-so-ever, and you do not need them.

So, what does the product rule tell us?

In plain English, it states that the derivative of the product of two
functions is equal to the derivative of the first times the second function
plus the derivative of the second times the first.

Or, to use f and g, it states:
(f*g)' = f'*g+g'*f

Notice my use of apostrophe as prime.

So, we also know that the product rule works for all points, and so if it
works for all points, it works for a single point, yes? that's a simple
deduction we can make.

So, what does h'(3) mean? It means, take the derivative of h with respect to
x, and state the answer for x =3.

So what is the derivative of h with respect to x?

Well, h(x) = f(x)*g(x)

And we know that the derivative of f(x)*g(x) is equal to
f'(x)*g(x)+g'(x)*f(x), right?

Now we put in three, so we know that h'(3) = f'(3)*g(3)+g'(3)*f(3)

Now, if only we knew what f(3), f'(3), g(3), and g'(3) were equal to? Oh
snap! We do, since we were given those four knowns, yes? so, plug them, do
the simple arithmetic, and you're done.

Hope that helps. Please make sure to really understand the product rule and
how it can be used.

Take care,
Sina

President, Prime Access Consulting, Inc.
Twitter: @SinaBahram
Company Website: http://www.pac.bz
Personal Website: http://www.sinabahram.com
Blog: http://blog.sinabahram.com

-----Original Message-----
From: Blindmath [mailto:blindmath-bounces at nfbnet.org] On Behalf Of Joseph
Lee via Blindmath
Sent: Saturday, October 18, 2014 5:07 PM
To: 'Sarah Jevnikar'; 'Blind Math list for those interested in mathematics'
Subject: Re: [Blindmath] Derivatives Help

Hi,
I don't think it's quotient rule, as the expression indicates that product
rule should be used. As Sarah said, what are f(x) and g(x) in this problem,
and does it ask that h(x) be derivative of F times G?
Cheers,
Joseph

-----Original Message-----
From: Blindmath [mailto:blindmath-bounces at nfbnet.org] On Behalf Of Sarah
Jevnikar via Blindmath
Sent: Saturday, October 18, 2014 2:01 PM
To: 'Jacques Chappell'; 'Blind Math list for those interested in
mathematics'
Subject: Re: [Blindmath] Derivatives Help

Do you have the original f(x) and g(x)? You'll have to differentiate first
with the quotient rule, then substitute 3 in.


-----Original Message-----
From: Blindmath [mailto:blindmath-bounces at nfbnet.org] On Behalf Of Jacques
Chappell via Blindmath
Sent: October-18-14 4:54 PM
To: 'Blind Math list for those interested in mathematics'
Subject: [Blindmath] Dirivitives Help

I am so lost on this problem and I feel like I can solve it, but I just
don't know where to begin or what to do. The problem is.

Find h'(3) when h(x)=f(x)*g(x)

g(3)=4

G'(3)=5

F(3)=9

f'(3)=8

 

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