How do I Apply the Product Rule in Calculus?

[Jake Minneman]

Hi I am 14 and attempting to learn calculus I have just proved product rule and am beginning examples of how it might work. Could anyone check, I will write my process.
$$y(x)=(12x^6)(7x^4+6)=$$
$$(12x^6)'(7x^4+6)+(12x^6)(7x^4+6)'=$$
$$(72x^5)(7x^4+6)+(12x^6)(28x^3)=$$
$$504x^9+432x^5+336x^9=$$
$$y(x)=840x^9+432x^5$$

 

[QuarkCharmer]

Hi I am 14 and attempting to learn calculus I have just proved product rule and am beginning examples of how it might work. Could anyone check, I will write my process.
$$y(x)=(12x^6)(7x^4+6)=$$
$$(12x^6)'(7x^4+6)+(12x^6)(7x^4+6)'=$$
$$(72x^5)(7x^4+6)+(12x^6)(28x^3)=$$
$$504x^9+432x^5+336x^9=$$
$$y(x)=840x^9+432x^5$$

Let's see, the product rule says that
$$(f*g)' = f'*g + f*g'$$

This looks good.
$$y(x)=(12x^6)(7x^4+6)=$$
$$(12x^6)'(7x^4+6)+(12x^6)(7x^4+6)'=$$

Then you differentiate, looks good.
$$(72x^5)(7x^4+6)+(12x^6)(28x^3)=$$

Now you simplify.
$$504x^9+432x^5+336x^9=$$
$$y(x)=840x^9+432x^5$$

Looks good so far, the only thing that I would do, is factor out a 24x^5 at the end there, but you are correct either way.

 

[Jake Minneman]

Thanks for the response

 

[Mark44]

You'll probably notice that your post was moved from the Precalc section to this one. This is the right place for Calculus homework problems.

 

[Jake Minneman]

Okay that's fine, LaTex is awesome.