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# C3 edexcel textbook chapter 8 differentiation watch

1. Hey,
This is like my first thread but got a quick question. It's on differentiation in C3.
Using the product rule, differentiate x(1+3x)^5.
Thanks for any help
2. (Original post by rainbowc4k3)
Hey,
This is like my first thread but got a quick question. It's on differentiation in C3.
Using the product rule, differentiate x(1+3x)^5.
Thanks for any help
What have you tried so far?
3. (Original post by rainbowc4k3)
Hey,
This is like my first thread but got a quick question. It's on differentiation in C3.
Using the product rule, differentiate x(1+3x)^5.
Thanks for any help
You know that,

Can you make a start from there?
4. (Original post by NotNotBatman)
What have you tried so far?
I've tried using the product rule and I've seen the markscheme:

let u=x v= (1+ 3x)^5

du/dx = 1 dv/dx= 5*3(1+3x)^4

(product rule) ====> dy/dx = u(dv/dx)+v(du/dx)
---> dy/dx = x*15(1+3x)^4 + (1+3x)^5 *1
I can do it up to here but then I get confused
=((1+3x)^4)(15x+1+3x)
=(1+3x)^4 *(1+18x)
How did they move the 15x into the other bracket and where does the power of 5 go?
5. (Original post by _gcx)
You know that,

Can you make a start from there?
I made a start, but I'm confused even though I've seen the MS
6. (Original post by rainbowc4k3)
I've tried using the product rule and I've seen the markscheme:

let u=x v= (1+ 3x)^5

du/dx = 1 dv/dx= 5*3(1+3x)^4

(product rule) ====> dy/dx = u(dv/dx)+v(du/dx)
---> dy/dx = x*15(1+3x)^4 + (1+3x)^5 *1
I can do it up to here but then I get confused
=((1+3x)^4)(15x+1+3x)
=(1+3x)^4 *(1+18x)
How did they move the 15x into the other bracket and where does the power of 5 go?
Expanding the brackets again,

Is it clearer now?
7. (Original post by rainbowc4k3)
I've tried using the product rule and I've seen the markscheme:

let u=x v= (1+ 3x)^5

du/dx = 1 dv/dx= 5*3(1+3x)^4

(product rule) ====> dy/dx = u(dv/dx)+v(du/dx)
---> dy/dx = x*15(1+3x)^4 + (1+3x)^5 *1
I can do it up to here but then I get confused
=((1+3x)^4)(15x+1+3x)
=(1+3x)^4 *(1+18x)
How did they move the 15x into the other bracket and where does the power of 5 go?
Factorise out (1+3x)^4.

because of the power rule.
Now factorise (1+3x)^4 and consider what each term is multiplied by.

or it might be easier to see if we let , then
Could you factorise that and then substitute a in terms of (1+3x)^4 at the end?
8. Oh wowwwwww...
I finally get it, thank you guys so much for the help it's literally so easy now that I see it.
Thanks guys much appreciated.
9. y = f(x)g(x), then [f(x)g(x)]’ = f’(x) g(x)+ f(x) g’(x)
for x(1+3x)^5

take f(x) as x

take g(x) as (1+3x)^5

then

f'(x) = 1
g'(x) = 15x(1+3x)^4

then just substitute into rule.

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