How do you differentiate y=(5x43x21)(5x2+3) using the product rule?

1 Answer
Jun 26, 2017

y'=150x5+120x38x

Explanation:

Take two functions.
E.g. f(x) and g(x).
The product of those functions is:
f(x)g(x)=(fg)(x)

When we talk about derivatives, the product rule can be used to write the derivative of said functions. Using the functions from before, it states:
(fg)'(x)=f'(x)g(x)+f(x)g'(x)

In your formula, we can see the "" symbol is between the brackets. So we could say something like:
(5x43x21)(5x2+3)=f(x)g(x).
f(x)=5x43x21
g(x)=5x2+3

(fg)'(x)=(5x43x21)'(5x2+3)+(5x43x21)(5x2+3)'
[Here we use the fact that we calculate each part seperately
and leave the constant, e.i (5x25x)'=(5x2)'(5x)'=5(x2)'5(x)']

=(20x36x)(5x2+3)+(5x43x21)(10x)
=100x5+60x3+30x318x50x5+30x3+10x
=150x5+120x38x