YAMT: Calculus (mat124), need help with derivatives (solved)

[Ricemarine]

Ok, so

f(x) = x^(2/3) * (x-4)^2

I'm trying to get f ' (x)...
So, using the product rule, I get...

f'(x) = 2/3x^(-1/3) * (x-4)^2 + x^(2/3) * 2 (x-4)

So to get it more simplified, I take out a 2/3x^(-1/3)

I SHOULD get (or so students say)

f ' (x) = 2/3x^(-1/3) [(x-4)^2 + 3x * (x-4)

Can someone help me on how they got to there?
I'm not really clear how the 3X got there...
Can you also help me with the second derivative? (optional...)

Thanks.

Edit: I get it.... the 2/3 would divide by 2 to become 3, therefore its x * 3 * (x-4)^2...
Yeah...

I'm trying not to confuse myself further .

 

[HaxorNubcake]

Originally posted by: Ricemarine
Ok, so

f(x) = x^(2/3) * (x-4)^2

I'm trying to get f ' (x)...
So, using the product rule, I get...

f'(x) = 2/3x^(-1/3) * (x-4)^2 + x^(2/3) * 2 (x-4)

So to get it more simplified, I take out a 2/3x^(-1/3)

I SHOULD get (or so students say)

f ' (x) = 2/3x^(-1/3) [(x-4)^2 + 3x * (x-4)

Can someone help me on how they got to there?
I'm not really clear how the 3X got there...
Can you also help me with the second derivative? (optional...)

Thanks.

they just factor. 2/(2/3) is 3

 

[3NF]

Originally posted by: Ricemarine
Ok, so

f(x) = x^(2/3) * (x-4)^2

I'm trying to get f ' (x)...
So, using the product rule, I get...

f'(x) = 2/3x^(-1/3) * (x-4)^2 + x^(2/3) * 2 (x-4)

So to get it more simplified, I take out a 2/3x^(-1/3)

I SHOULD get (or so students say)

f ' (x) = 2/3x^(-1/3) [(x-4)^2 + 3x * (x-4)

Can someone help me on how they got to there?
I'm not really clear how the 3X got there...
Can you also help me with the second derivative? (optional...)

Thanks.

This looks right,

2/3x^(-1/3) * (x-4)^2 + x^(2/3) * 2 (x-4)

Taking out the 2/3x^(-1/3) will give you the 3x

2/3x^(-1/3) * [(x-4)^2 + 3x * (x-4)] ... multiply it out, and you'll see you have what you started with

 

[FleshLight]

edit: NM. As for the 2nd derivative, it's really ugly. You have to use chainrule + product rule a few times.

 

[Ricemarine]

Originally posted by: HaxorNubcake

Originally posted by: Ricemarine
Ok, so

f(x) = x^(2/3) * (x-4)^2

I'm trying to get f ' (x)...
So, using the product rule, I get...

f'(x) = 2/3x^(-1/3) * (x-4)^2 + x^(2/3) * 2 (x-4)

So to get it more simplified, I take out a 2/3x^(-1/3)

I SHOULD get (or so students say)

f ' (x) = 2/3x^(-1/3) [(x-4)^2 + 3x * (x-4)

Can someone help me on how they got to there?
I'm not really clear how the 3X got there...
Can you also help me with the second derivative? (optional...)

Thanks.

they just factor. 2/(2/3) is 3

But where exactly did the 2/3 come from?... Or why did the 2 be divided by the 2/3?...

 

[FleshLight]

Originally posted by: Ricemarine

Originally posted by: HaxorNubcake

Originally posted by: Ricemarine
Ok, so

f(x) = x^(2/3) * (x-4)^2

I'm trying to get f ' (x)...
So, using the product rule, I get...

f'(x) = 2/3x^(-1/3) * (x-4)^2 + x^(2/3) * 2 (x-4)

So to get it more simplified, I take out a 2/3x^(-1/3)

I SHOULD get (or so students say)

f ' (x) = 2/3x^(-1/3) [(x-4)^2 + 3x * (x-4)

Can someone help me on how they got to there?
I'm not really clear how the 3X got there...
Can you also help me with the second derivative? (optional...)

Thanks.

they just factor. 2/(2/3) is 3

But where exactly did the 2/3 come from?... Or why did the 2 be divided by the 2/3?...

2/3 is from the derivative of x^2/3. You then divide f'(x) by 2/3 to get rid of it.

 

[SoLiDus88]

man wait till you get to integrals...

 

[AFB]

*pulls out trusty ti-89 because he's a lazy american*

*discovers it's in his car*

*works by hand*

f'(x) = (2/3)-X^(-1/3) *(x-4)^2 + x^(2/3) * (2) *(x-4)
f"(x) = (-2/9)X^(-4/3) *(x-4)^2 + (2)(x-4)(2/3)(x^(-1/3)) + 2(2/3x^(-1/3) * (x-4) + x^(2/3))

That's pretty nasty but it's certainly doable. I'm sure I missed a coefficient or something somewhere.