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

According to the product rule,
# ( f ⋅ g ) ′ = f ′ ⋅ g + f ⋅ g ′#
This just means that when you differentiate a product, you do derivative of the first, leave second alone, plus derivative of the second, leave the first alone.

So the first would be #(x^3 − 3x)# and the second would be #(2x^2 + 3x + 5)#.

Okay, now the derivative of the first is #3x^2-3#, times the second is #(3x^2-3)*(2x^2 + 3x + 5)#.

The derivative of the second is #(2*2x+3+0)#, or just #(4x+3)#.
Multiply it by the first and get #(x^3 − 3x)*(4x+3)#.

Add both portions together now: #(3x^2-3)*(2x^2 + 3x + 5) + (x^3 − 3x)*(4x+3)#

If you multiply it all out and simplify, you should get #10x^4+12x^3-3x^2-18x-15#.

The product rule states that for a function, #f# such that;

#f(x) = g(x)h(x)#
#d/dx f(x) = g'(x)h(x) + g(x)h'(x)#

The function #f# is given as #f(x) = (x^3-3x)(2x^2+3x+5)#, which we can split into the product of two functions #g# and #h#, where;

#g(x) = x^3 - 3x#
#h(x) = 2x^2+3x+5#

By applying the power rule, we see that;

#g'(x) = 3x^2 - 3#
#h'(x)=4x+3#

Plugging #g#, #g'#, #h#, and #h'# into our power rule function we get;

#d/dx f(x) = (3x^2 - 3)(2x^2+3x+5) + (x^3 - 3x)(4x+3)#

#d/dx f(x)=6x^4+9x^3+15x^2-6x^2-9x-15+4x^4+3x^3-12x^2-9x#

#d/dx f(x)=10x^4+12x^3-3x^2-18x-15#