The product rule (1 Viewer)

[Twickel]

Hi

I get his after using the product rule
60x(3x-2)^4+ 4(3x-2)^5

How do I simplify it to get 8(3x-2)^4(9x-1)

I have no idea what to do every question I cant simplify, can someone please post the steps on what to do please.

This is what I do 60x/10 = 6x then 6(3x-2)^4 { 6x+(3x-2)}
6(3x-2)^4(9x-2)

What am I doing wrong? note ^4 is to the pwer of 4 .

 

[midifile]

Hi

I get his after using the product rule
60x(3x-2)^4+ 4(3x-2)^5

How do I simplify it to get 8(3x-2)^4(9x-1)

I have no idea what to do every question I cant simplify, can someone please post the steps on what to do please.

This is what I do 60x/10 = 6x then 6(3x-2)^4 { 6x+(3x-2)}
6(3x-2)^4(9x-2)

What am I doing wrong? note ^4 is to the pwer of 4 .

Is there meant to be an x after the 60? Because the qu works if there is not

You have 60(3x-2)^4+ 4(3x-2)^5. The common factor of the 2 terms is 4(3x-2)^4
therefore, 60(3x-2)^4+ 4(3x-2)^5 = 4(3x-2)^4(15x+(3x-2))
=4(3x-2)^4(18x-2)
=4(3x-2)^4 x 2(9x-1)
=8(3x-2)^4(9x-1)

 

[Twickel]

Thanks so much, I must go now but if I have any other problems related to his topic can I post it here?

3x^2(x+1)^2 + 2x(x+1)^2

show please.

 

[midifile]

Thanks so much, I must go now but if I have any other problems related to his topic can I post it here?

3x^2(x+1)^2 + 2x(x+1)^2

show please.

Yea sure

Common factor is x(x+1)^2
Therefore 3x^2(x+1)^2 + 2x(x+1)^2 = x(x+1)^2(3x+2)

 

[Aplus]

Thanks so much, I must go now but if I have any other problems related to his topic can I post it here?

3x^2(x+1)^2 + 2x(x+1)^2

show please.

You would probably get more responses, were this moved to the Mathematics forum.

 

[kurt.physics]

Yea sure

Common factor is x(x+1)²
Therefore 3x²(x+1)² + 2x(x+1)² = x²(x+1)²(3x+2)

im sure this is wrong, but i could be wrong, it is only x, here is my work.

3x²(x+1)² + 2x(x+1)²

common factor is x(x+1)²

Let u = x(x+1)²

then it becomes

3xu +2u

= u(3x + 2)

subbing back in we get

x(3x+2)(x+1)²

 

[Twickel]

Ok

Tell me if this is correct 3x^4(4-x)^3

u= 3x^4 v= (4-x)^3
du/dx = 12x^3 dv/dx= -3

-3 ( 3x^4) = -9x^4 12x^3( 4-x)^3 = 12x^3(4-x)^3

Now I get -9x^4+ 12x^3(4-x)^3
HCF= -3x^3

u= -3x^3
3xu+-4u

GRRR ITS WRONG.

I cant belive it I was doing fine in this topic till this came about,

Sorry for my stupidity

I diffed v wrong. ok let me try once i diff it right.

ok HCF= -3x^3(4-x)^2???

 

[midifile]

im sure this is wrong, but i could be wrong, it is only x, here is my work.

3x²(x+1)² + 2x(x+1)²

common factor is x(x+1)²

Let u = x(x+1)²

then it becomes

3xu +2u

= u(3x + 2)

subbing back in we get

x(3x+2)(x+1)²

typo

= now fixed

 

[midifile]

u= 3x^4 v= (4-x)^3
du/dx = 12x^3 dv/dx= -3

dv/dx = -3(4-x)^2

Try following that through

 

[Twickel]

The answer is 3x^3(4-x)^2(-7x+16)
I got 3x^3(4-x)^2 ( -3x+4)

If the hcf= 3x^3( 4-x)^2

then using kurts u method I get --3xu+4u= ( -3x+4)

 

[midifile]

The answer is 3x^3(4-x)^2(-7x+16)
I got 3x^3(4-x)^2 ( -4+3x)

If the hcf= 3x^3( 4-x)^2

then using kurts u method I get --3xu+4u= ( -3x+4)

u=3x^4
u' = 12x^3
v=(4-x)^3
v' = -3(4-x)^2

12x^3(4-x)^3 - 9x^4(4-x)^2 = 12x^3(4-x)^3 - 9x^4(4-x)^2
= 3x^3(4-x)^2(4(4-x) -3x)
= 3x^3(4-x)^2(16 - 4x -3x)
= 3x^3(4-x)^2(- 7x+16)

You dont really have to take out the common factor like kurt did unless all the numbers get confusing.

 

[Twickel]

Ok, so when I get to the part were I need to simplify the equation what must I do? Look for the HCF then ................................ so in that question your saying the common factor to those is?


I think my problem is picking out the Common factor, I really cant pick it out

 

[midifile]

Ok, so when I get to the part were I need to simplify the equation what must I do? Look for the HCF then ................................ so in that question your saying the common factor to those is?


I think my problem is picking out the Common factor, I really cant pick it out

Okay you look for the highest common factor and take it out. For example it the last question both terms had numbers, x's and (4-x)'s.

So find the biggest number that goes into both numbers (12 and 9) which is 3, the highest power of x that goes into both of the x's (x^3 and x^4) which is x^3 and the highest power of (4-x) that does into both the (4-x)'s which is (4-x)^2. Put them together and that is your highest common factor. So in the last question it was 3x^3(4-x)^2.

Then you just work out what is left over, and put that in the brackets next to the HCF. Then simplify this if possible.

 

[Twickel]

What happens to the - 9x^4 you just sub the answer of the dividing part for - 9x^4?

e.g if it was 12/4=3 the - 9x^4 become a 3?

 

[midifile]

What happens to the - 9x^4 you just sub the answer of the dividing part for - 9x^4?

e.g if it was 12/4=3 the - 9x^4 become a 3?

Because you have taken out a 3 as part of your common factor, you divide the -9 by 3 which leaves you with -3.

If you are getting confused with all the x's and stuff think of a simple question like: factorise 12 + 9x

Common factor is 3, so 12 - 9x = 3(4-3x).

Thats basically what you are doing in those question, but youve just got more x's, powers etc.

 

[Twickel]

Oh God the next part of this topic is so much easier then this.

I think the book madea mistake in this question because the rest Im getting right. x^3/x^2-4 =??
using vxdu/dx-uxdv/v^2

 

[Aplus]

Oh God the next part of this topic is so much easier then this.

I think the book madea mistake in this question because the rest Im getting right. x^3/x^2-4 =??
using vxdu/dx-uxdv/v^2

f(x) = x³/x² - 4
f'(x) = [(3x²)(x² - 4) - (2x)(x³)]/(x² - 4)²
= 3x⁴ - 12x² - 2x⁴/(x² - 4)²
= x⁴ - 12x²/(x² - 4)²