How do you differentiate #y=(5x^4-3x^2-1)(-5x^2+3)# using the product rule?

1 Answer
Jun 26, 2017

Answer:

#y' = -150x^5+120x^3-8x#

Explanation:

Take two functions.
E.g. #f(x)# and #g(x)#.
The product of those functions is:
#f(x)*g(x) = (f*g)(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:
#(f * g)'(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:
#(5x^4-3x^2-1) * (-5x^2+3) = f(x) * g(x)#.
#f(x) = 5x^4-3x^2-1#
#g(x) = -5x^2+3#

#(f * g)'(x) = (5x^4-3x^2-1)' * (-5x^2+3) + (5x^4-3x^2-1) * (-5x^2+3)'#
[Here we use the fact that we calculate each part seperately
and leave the constant, e.i #(5x^2-5x)' = (5x^2)'-(5x)' = 5(x^2)' - 5(x)'#]

#= (20x^3-6x) * (-5x^2+3) + (5x^4-3x^2-1) * (-10x)#
#= -100x^5+60x^3+30x^3-18x -50x^5+30x^3+10x#
#= -150x^5+120x^3-8x#