Help Differentiate? (1 Viewer)

[Zokunu]

Use the product rule to differentiate.

y= x^2 (x^3 - 2x + 5)

y = (7x^3 -1)(2x^2 + 6x + 5)

THX

 

[EpikHigh]

Use the product rule to differentiate.

y= x^2 (x^3 - 2x + 5)

THX

So we know the product rule is vu' + uv' right?

u = x^2
u'= 2x

v = x^3-2x+5
v' = 3x^2-2

therefore dy/dx = 2x(x^3-2x+5) + x^2(3x^2-2) expand then simplify.

 

[Zokunu]

So we know the product rule is vu' + uv' right?

u = x^2
u'= 2x

v = x^3-2x+5
v' = 3x^2-2

therefore dy/dx = 2x(x^3-2x+5) + x^2(3x^2-2) expand then simplify.

mhm thx man

 

[Smile12345]

2nd Question - Same as EpikHigh's explanation.

u=7x^3 -1
u'=21x^2

v= 2x^2 + 6x + 5
v'= 4x + 6

Therefore dy/dx = (2x^2 + 6x + 5).(21x^2) + (7x^3-1).(4x + 6)
which is the same as 21x^2 (2x^2 + 6x + 5) + 4x+6(7x^3 -1)

Then expand and simplify.

Trust I've got it right.

 

[Zokunu]

2nd Question - Same as EpikHigh's explanation.

u=7x^3 -1
u'=21x^2

v= 2x^2 + 6x + 5
v'= 4x + 6

Therefore dy/dx = (2x^2 + 6x + 5).(21x^2) + (7x^3-1).(4x + 6)
which is the same as 21x^2 (2x^2 + 6x + 5) + 4x+6(7x^3 -1)

Then expand and simplify.

Trust I've got it right.

Ye thx, i did that one already . Can you guys help me out with this? I did it, but what i got is a long as...equation which i think is wrong.

y = (x^2+5x)(4x -11)^6

 

[EpikHigh]

Ye thx, i did that one already . Can you guys help me out with this? I did it, but what i got is a long as...equation which i think is wrong.

y = (x^2+5x)(4x -11)^6

This is very similar, It's a combination of both the product rule and chain rule, that is with the 2nd product (4x-11)^6

So this is the exact same, except when you differentiate v you use chain rule

v = (4x-11)^6
dv/dx = 6(4x-11)^5 multiplied by 4 (the derivative of whats inside the brackets

which is 24(4x-11)^5

the rest I think you should be able to do.

 

[Zokunu]

This is very similar, It's a combination of both the product rule and chain rule, that is with the 2nd product (4x-11)^6

So this is the exact same, except when you differentiate v you use chain rule

v = (4x-11)^6
dv/dx = 6(4x-11)^5 multiplied by 4 (the derivative of whats inside the brackets

which is 24(4x-11)^5

the rest I think you should be able to do.

I did what you said. I got this -> (2x+5)(4x-11)^6 + 24(x^2 + 5x)(4x-11)^5

Ok, how am I supposed to factorise this?
Is it (4x-11)^5 {(2x+5) + 24 (x^2 + 5x)(4x-11)?

Thx man

 

[SharkeyBoy]

I did what you said. I got this -> (2x+5)(4x-11)^6 + 24(x^2 + 5x)(4x-11)^5

Ok, how am I supposed to factorise this?
Is it (4x-11)^5 {(2x+5) + 24 (x^2 + 5x)(4x-11)?

Thx man

Close, it's
(4x-11)^5 {(2x+5)(4x-11) + 24 (x^2 + 5x)}

 

[Zokunu]

Close, it's
(4x-11)^5 {(2x+5)(4x-11) + 24 (x^2 + 5x)}

Bump, Ye thx man

Can you guys help me with this question?

y = (3x+5)^6 √x

 

[asianese]

by the product rule.

 

[Zokunu]

by the product rule.

Sorry man, can you explain a bit further?

 

[cookiez69]

Sorry man, can you explain a bit further?

.
All you do is use the product rule and when you differentiate for u' and v', you must use the product rule as well. Try it for yourself first, but if you need the working out tell me.

 

[Zokunu]

.
All you do is use the product rule and when you differentiate for u' and v', you must use the product rule as well. Try it for yourself first, but if you need the working out tell me.

A bit confusing but ye. I got it now, thx

 

[koreafantasy]

Screw the vuuv method, I find it really hard to use in exam situations. Try this instead-
Write out the two functions you are multiplying with brackets twice, and add a plus in the middle. Add a dash to the left one first, then on the other side, a dash to the right one. say the function you are trying to differentiate is (2x+1)(3x+1).
You write-
(2x+1)'(3x+1) + (2x+1)(3x+1)'
= 2(3x+1) + 3(2x+1)
= Easy algebra
This way, you can really easily see what you are doing, rather than trying to figure out what v and u are, and trying to differentiate them separately.