Year 11 Double variable derivatives

[DJ-Smiles]

URGENT !Year 11 Double variable derivatives

I am having trouble with this question it is derivatives. Previously I have been able to complete these with no trouble but am a little confused with how start this one:
y= a^2(3x+5)^3.

I don't know whether to use the product rule and just leave it with two unknown variables or try and find a value for both a and x.

I tried this :
u= a^2
u'= 2a
v=(3x+5)^3
v= 9(3x+5)^2
y'= v'u +u'v
= 9a^2(3x+5)^2 + 2a(3x+5)^3
=a(3x+5)^2[2(3x+5) +9a)]
=a(3x+5)^2(6x+5+9a)

Help me please

 

[Inertigratus]

Is 'a' a variable or a constant?

 

[DJ-Smiles]

I have no idea :( I am so lost and i need it for tomorrow :/ it's just basic differential calculus. what would you do?

 

[Inertigratus]

I think it's a constant, meaning that y = f(x) :).
Then you just treat 'a' as a number.
Think you can do it from here?

 

[DJ-Smiles]

Ok um can you guide me through it ?

 

[Inertigratus]

Well, since a^2 is just a number, you don't have to do anything with it.
So what's the derivative of (3x+5)^3 ?

 

[DJ-Smiles]

so it would be 9(3x+5)^2? is the working i did at the top correct ? btw thanks heaps for your help

 

[Inertigratus]

Sure, no problem. Yeah it's right :) then that times a^2

 

[DJ-Smiles]

does that mean I don't need to use the derivative of a^2

 

[Inertigratus]

Yes, because you're taking the derivative with respect to x. 'a' isn't a function of x, so the derivative of 'a' is simply 0.
So the whole derivative is 9a^2(3x+5)^2.

 

[DJ-Smiles]

Ohh thanks Lol I understand now! :)