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

2 Answers
Apr 13, 2018

The answer is (3x^2-3)*(2x^2 + 3x + 5) + (x^3 − 3x)*(4x+3), which simplifies to 10x^4+12x^3-3x^2-18x-15.

Explanation:

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.

Apr 13, 2018

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

Explanation:

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