Differential calc..product rule

**playpwnsu** · November 12th 2009, 07:35 PM

1. Let f(x) = e^nx, where n is a positive integer. Find f'(x) (Hint: use the Product Rule)

Okay, it says to use the product rule, thats d/dx = f'g +g'f
I put it as

f'(x) = (ne^nx)g(x) + g'(x)(e^nx)

Im not sure if thats right...because it does not give the g(x), it is there someway to figure it out?

2. Let f1, f2, f3, and f4 be differentiable functions

(a) Find d/dx [f1(x)f2(x)f3(x)]

okay so what i did was, I applied the quotient rule and wrote the above as
= f'g -g'f/g^2
= f1f2/(1/f3)
= [f1f2]'[1/f3] - [f'3/f3^2][f1f2]/(1/f3)^2

Is that correct????

(b) Find d/dx [f1(x)f2(x)f3(x)f4(x)]

I basically did what I have done above, and used the product rule

and got
f'(x) = (f1f2)'(f3f4) + (f3f4)'(f1f2)

Anyhelp would be appreciated, thanks guys

**redsoxfan325** · November 12th 2009, 07:42 PM

Originally Posted by playpwnsu

1. Let f(x) = e^nx, where n is a positive integer. Find f'(x) (Hint: use the Product Rule)

Okay, it says to use the product rule, thats d/dx = f'g +g'f
I put it as

f'(x) = (ne^nx)g(x) + g'(x)(e^nx)

Im not sure if thats right...because it does not give the g(x), it is there someway to figure it out?

2. Let f1, f2, f3, and f4 be differentiable functions

(a) Find d/dx [f1(x)f2(x)f3(x)]

okay so what i did was, I applied the quotient rule and wrote the above as
= f'g -g'f/g^2
= f1f2/(1/f3)
= [f1f2]'[1/f3] - [f'3/f3^2][f1f2]/(1/f3)^2

Is that correct????

(b) Find d/dx [f1(x)f2(x)f3(x)f4(x)]

I basically did what I have done above, and used the product rule

and got
f'(x) = (f1f2)'(f3f4) + (f3f4)'(f1f2)

Anyhelp would be appreciated, thanks guys

Are you sure you don't mean the chain rule?

$\frac{d}{dx}[e^{nx}]=e^{nx}\cdot\frac{d}{dx}[nx]=ne^{nx}$

If you want to use the product rule, you're going to have to treat $e^{nx}$ as $\underbrace{e^x\cdot e^x\cdot...\cdot e^x}_{n~\text{times}}$ and use induction to show that $\frac{d}{dx}[e^{nx}]=ne^{nx}$ .

For the second one, why are you using the quotient rule for the product of three functions? Just use the product rule twice.

**playpwnsu** · November 12th 2009, 07:48 PM

Yes, for the first question, we have not learned the chain rule yet, but yes I am aware of it, but in this case our prof wants us to use the product rule, so i guess we have to use that.

What do you mean to use induction?

Also, for number 2, if i apply the product rule twice to the 3 functions, do i need to apply it twice for the 4 functions too?

Thanks

**redsoxfan325** · November 12th 2009, 08:14 PM

You would need to apply it three times for four functions.

------------

Induction

1.) Prove it's true for a base base (i.e. $n=1$ ).
2.) Assume it's true for $n=k$ .
3.) Prove it's true for $n=k+1$ .

For this problem,

1.) $\frac{d}{dx}[e^x]=e^x$ . Done

2.) Assume $\frac{d}{dx}[e^{kx}]=ke^{kx}$ . Done

3.) Prove that $\frac{d}{dx}[e^{(k+1)x}]=(k+1)e^{(k+1)x}$ .

This is where you use the product rule. Note that

$e^{(k+1)x}=e^{kx+x}=e^{kx}\cdot e^x$

Now use the product rule and the assumption from (2) to show that $\frac{d}{dx}[e^{kx}\cdot e^x]=(k+1)e^{(k+1)x}$ .

**playpwnsu** · November 12th 2009, 09:14 PM

Thanks for the help with those two questions

I have 2 more questions that I need help with if you dont mind,

Okay
1. Let f(x) = x^2/e^2x. Solve f''(x) = 0

What I did was use the quotient rule : (f'g - g'f)/g^2
= x^2/e^2x
= {[2x][e^2x] - [2e^2x][x^2]}/(e^2x)^2
= [e^2x(2x - 2x^2)]/(e^2x)^2
f'(x) = [2x-2x^2]/e^2x
Then I took the derivative of f'(x)
= {[2-4x][e^2x] - [2e^2x][2x-2x^2]}/(e^2x)^2
= [4x^2-8x+2]/e^2x
=2[2x^2-4x+1]/e^2x

I cannot get it to = 0?? what am I doing wrong?

2. Before proving the Product Rule as we know it, Gottfried Leibniz reportedly assumed that
d/dx [f(x)g(x)] = [d/dx f(x)][d/dx g(x)]
Find an example of functions f and g such that this equation holds

Well, I read in another post about this problem to use a polynomial for f, and use 1/x for g, and then apply then product rule?
but I cannot seem to get f'g'....and also some other method required the use of integration..but we have not learned integration yet...so is there anyway I can do this problem with what I have learned so far??

Thanks, any help would be appreciated...

**redsoxfan325** · November 12th 2009, 09:25 PM

Originally Posted by playpwnsu

Thanks for the help with those two questions

I have 2 more questions that I need help with if you dont mind,

Okay
1. Let f(x) = x^2/e^2x. Solve f''(x) = 0

What I did was use the quotient rule : (f'g - g'f)/g^2
= x^2/e^2x
= {[2x][e^2x] - [2e^2x][x^2]}/(e^2x)^2
= [e^2x(2x - 2x^2)]/(e^2x)^2
f'(x) = [2x-2x^2]/e^2x
Then I took the derivative of f'(x)
= {[2-4x][e^2x] - [2e^2x][2x-2x^2]}/(e^2x)^2
= [4x^2-8x+2]/e^2x
=2[2x^2-4x+1]/e^2x

I cannot get it to = 0?? what am I doing wrong?

The derivative is correct. To make it equal to zero, you need the numerator equal to zero. Just use the quadratic formula on $2x^2-4x+1=0$ .

Originally Posted by playpwnsu

2. Before proving the Product Rule as we know it, Gottfried Leibniz reportedly assumed that
d/dx [f(x)g(x)] = [d/dx f(x)][d/dx g(x)]
Find an example of functions f and g such that this equation holds

Well, I read in another post about this problem to use a polynomial for f, and use 1/x for g, and then apply then product rule?
but I cannot seem to get f'g'....and also some other method required the use of integration..but we have not learned integration yet...so is there anyway I can do this problem with what I have learned so far??

Thanks, any help would be appreciated...

How about $f(x)=g(x)=e^{2x}$ ?

$f(x)g(x)=e^{4x}$ , so the derivative is $4e^{4x}$
$f'(x)g'(x)=2e^{2x}\cdot2e^{2x}=4e^{4x}$

**playpwnsu** · November 12th 2009, 09:29 PM

well, actually i worked on the 2nd question again and, If I use f = x^2
g = 1/x

This is what I did
d/dx[f(x)g(x)] = f'g'
= [2x][[x^-1] + [-x^-2][x^2]
= x^-1[2x-x^-1*x^2]
= x^-1[2x-x]
f'g'= x^-1[x]

This is different in that, if you take f = x^2, then f' = 2x, and g = 1/x and g' = -x^-2, so the answer should be f'g' = -x^-2[2x], but thats not what i got

**redsoxfan325** · November 12th 2009, 09:32 PM

Originally Posted by playpwnsu

well, actually i worked on the 2nd question again and, If I use f = x^2
g = 1/x

This is what I did
d/dx[f(x)g(x)] = f'g'
= [2x][[x^-1] + [-x^-2][x^2]
= x^-1[2x-x^-1*x^2]
= x^-1[2x-x]
f'g'= x^-1[x]

This is different in that, if you take f = x^2, then f' = 2x, and g = 1/x and g' = -x^-2, so the answer should be f'g' = -x^-2[2x], but thats not what i got

If $f(x)=x^2$ and $g(x)=\frac{1}{x}$ , then $f(x)g(x)=x$ so $\frac{d}{dx}[fg]=1$

Using the incorrect method, $f'(x)=2x$ and $g'(x)=-\frac{1}{x^2}$ , so $f'(x)g'(x)=-\frac{2}{x}$

So this doesn't work as an example.

**playpwnsu** · November 12th 2009, 09:50 PM

I used the quadratic formula for 2x^2-4x+1=0

{-b(+_)sq[b2-4ac]}/2a
a= 2 b = -4 c= 1

={-(-4)+_sq[(-4)^2 -4(2)(1)]}/2(2)
=1+_sq[8]

is it suppose to equate to 0?

**redsoxfan325** · November 12th 2009, 10:09 PM

Yes.

$x=\frac{4\pm2\sqrt{2}}{4}=1\pm\frac{\sqrt{2}}{2}$

**playpwnsu** · November 12th 2009, 10:19 PM

wow oh yeah..thanks hehe

okay hmm, i have one last question, that I'm really not sure what to do...

1. It may be shown that the infinite sum
1+1/2+1/2^2+1/2^3.....
is equal to 2; and that, in general, for c E(can't do that symbol....) (-1,1),
1+c+c^2+c^3+....=1/1-c
Using this fact, prove that for c E (-1,1), c+2c^2+3c^3+4c^4+....=c/(1-c)^2

...thanks in advance, if you cannot help me, perhaps give me some hint about how this question works so I can get started, cheers

**redsoxfan325** · November 12th 2009, 10:23 PM

Originally Posted by playpwnsu

wow oh yeah..thanks hehe

okay hmm, i have one last question, that I'm really not sure what to do...

1. It may be shown that the infinite sum
1+1/2+1/2^2+1/2^3.....
is equal to 2; and that, in general, for c E(can't do that symbol....) (-1,1),
1+c+c^2+c^3+....=1/1-c
Using this fact, prove that for c E (-1,1), c+2c^2+3c^3+4c^4+....=c/(1-c)^2

...thanks in advance, if you cannot help me, perhaps give me some hint about how this question works so I can get started, cheers

$\sum_{n=0}^{\infty}c^n=\frac{1}{1-c}$

Differentiate both sides w.r.t. $c$ . Then multiply both sides by $c$ .

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