Differentiatio help please.....

Find dy/dx of y=(2x+3)(x+5)^2
Okay, so I'm using product rule and this is what i've done so far:

(2x+3)(x+5)^2
2 [x+5]^2 + 2x+3 (2) (x+5) (1)
=2 [x+5]^2 + 2*2x+3 (x+5)

What do I do next, show the steps please?

There are a number of ways you could do this. You could expand the 3 brackets and then it'd just be straight differentiation of a polynomial. You could use the product rule

$\frac{d}{dx}uv = \frac{du}{dx}v + u\frac{dv}{dx}$

Where

$u = 2x+3$

and

$v = (x+5)^2$

Or you could do the product rule of 3 functions.

$\frac{d}{dx}uvw = \frac{du}{dx}vw + u\frac{dv}{dx}w + uv\frac{dw}{dx}$

Where

$u = 2x+3$

and

$v = w = x+5$


A lot of people seem to forget to apply the chain rule, so the best method to ensure you do it correctly would be either to multiply it all out, or the last method, product rule of 3 functions.

Could you rewrite what you've got up to? What you have done seems to be correct, although don't forget you're multiplying the 2 from the $(x+5)^2$ with the entire function, not just the 2x ( you didn't put brackets around it).

2(x+5)^2 + 2(2x+3)(x+5)

2(x+5)[x+5+2x+3]

2(x+5)(3x+8)

How did you cancel out? what bits cancelled out each other? Thats the hardest part for me :|

Okay thanks And yeah the simplifying but is the difficult bit for me, I just dont no when and what to cancel out :|
But thanks for ur help

factorise 2(x+5) from both terms, this leaves you with x+5 in the first term and 2x+3 in the second. x+5+2x+3 = 3x+8