Multiplying Coefficients of different variables ?

[Hennessy]

Hi All, first post here. Currently finished the first year of my physics degree, with no background maths knowledge so everything's new. ( i have passed all coursework and exams)

I'm doing practice over the summer on key topics like partial derivatives, rules of differentiation, integration, multivariable calculus, ODE , first and second order differential equations, linear algebra etc.

(wrt = with respect to)

I made a function of the following:

$$f(x, y) = 3x^2y^4 + 6y + 2yx + 3x$$
differentiating wrt y i obtained :
$$4y^33x^2 + 6 + 2x$$
I put this into a differentiation calculator and it gave me the following result for the derivative of this function wrt y :
$$12x^2y^3 + 2x + 6$$
I noticed immediately that the coefficient 4 associated with the y variable has been multiplied by 3 which is the coefficient in front of the x variable. I haven't done differentiation in a while. So as a question is it okay to generally multiple coefficients if there's a part of the function where there are 2 coefficients , because I'm aware effectively it is saying $$(4y^3)(3x^2)$$ which would give $$12x^2y^3$$
Any help would be appreciated, looking forward to being on the forums a lot over the summer

 

[stapel]

Hi All, first post here. Currently finished the first year of my physics degree, with no background maths knowledge so everything's new. ( i have passed all coursework and exams)

I'm doing practice over the summer on key topics like partial derivatives, rules of differentiation, integration, multivariable calculus, ODE , first and second order differential equations, linear algebra etc.

(wrt = with respect to)

I made a function of the following:

$$f(x, y) = 3x^2y^4 + 6y + 2yx + 3x$$
differentiating wrt y i obtained :
$$4y^33x^2 + 6 + 2x$$
I put this into a differentiation calculator and it gave me the following result for the derivative of this function wrt y :
$$12x^2y^3 + 2x + 6$$
I noticed immediately that the coefficient 4 associated with the y variable has been multiplied by 3 which is the coefficient in front of the x variable. I haven't done differentiation in a while. So as a question is it okay to generally multiple coefficients if there's a part of the function where there are 2 coefficients , because I'm aware effectively it is saying $$(4y^3)(3x^2)$$ which would give $$12x^2y^3$$
Any help would be appreciated, looking forward to being on the forums a lot over the summer

Yes, you can simplify by completing the multiplication.

 

[Dr.Peterson]

I noticed immediately that the coefficient 4 associated with the y variable has been multiplied by 3 which is the coefficient in front of the x variable. I haven't done differentiation in a while. So as a question is it okay to generally multiply coefficients if there's a part of the function where there are 2 coefficients , because I'm aware effectively it is saying $$(4y^3)(3x^2)$$ which would give $$12x^2y^3$$

Yes, you can do that. We can always apply the commutative and associative properties to simplify a product, and it's usually a good idea to do so.

My question for you is, is there any particular reason you thought you couldn't, or shouldn't? That may help us see where, if at all, you might need correction.

 

[Steven G]

Variables are place holders for numbers. When you have to multiply a few numbers together (note that some of thse numbers may be variables) the order that you multiply in doesn't matter. So yes, you can multiply the 3 and 4 together!
In computing 25*7*4 there is no need to run and get a calculator as that will definitely take too long. Just multiply 25 and 4, get 100 and then multiply that result by 7 to get 700/

 

[Hennessy]

Yes, you can do that. We can always apply the commutative and associative properties to simplify a product, and it's usually a good idea to do so.

My question for you is, is there any particular reason you thought you couldn't, or shouldn't? That may help us see where, if at all, you might need correction.

Hello thank you for your timely response. The reason why i thought i couldn't was because the coefficients were associated with different variables , Because the coefficients are associated with different variables i thought that they should be left alone ? hopefully that gives you something! aha

 

[Dr.Peterson]

Hello thank you for your timely response. The reason why i thought i couldn't was because the coefficients were associated with different variables , Because the coefficients are associated with different variables i thought that they should be left alone ? hopefully that gives you something! aha

Thanks.

Each term is just a product (of numbers and variables); the numbers are not permanently associated with individual variables, even though that may be where they came from.

Once you have written a product, you can (and should) apply the commutative and associative properties to simplify the product, as I mentioned. They allow you to change the order from $(4y^3)(3x^2)$ to $(4)(3)y^3x^2$, and then do any multiplication first.

So we can write it as $12y^3x^2$. And then we commonly write the variables in alphabetical order, though that isn't necessary; so I'd usually write it as $12x^2y^3$, as they showed.

Similarly, if the term were $(4x^3)(3x^2)$, you can rearrange to $(4)(3)(x^3x^2)=12x^5$, multiplying the two powers of x together.

 

[Hennessy]

Appreciate all the kindful help and knowledgeable responses from people on here, thank you for your help!

- Hennessy