What is the derivative of #f(x) = (2x - 3)^4 (x^2 + x + 1)^5#?

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Jake L. Share
Sep 9, 2016

Answer:

#f'(x) = (2x-3)^3(x^2 + x + 1)^4(28x^2-12x-7)#

Explanation:

Looking at the equation,
#f(x) = (2x-3)^4(x^2+x+1)^5#

we first notice a couple patterns.
1. The function is a product of two terms
2. Each of the terms is a term with an exponent .

Since the function is a product of two terms, we know that we have to use the Product Rule to find the first derivative. The Product Rule states,

For an equation,
#f(x) = g(x) * h(x)#
#f'(x) = g'(x)h(x) + g(x)h'(x)#

In our case,
#f(x) = g(x) * h(x)#
where,
#g(x) = (2x-3)^4#
#h(x) = (x^2+x+1)^5#

Now we need to calculate g'(x) and h'(x) for the product rule. For this, we need the Chain Rule. It states,
If #F(x) = f(g(x))#,
#F'(x) = f'(g(x))*g'(x)#

so, in our specific case,
#g(x) = (2x-3)^4#
#g'(x) = 4(2x-3)^3 * (2)#
#h(x) = (x^2+x+1)^5#
#h'(x) = 5(x^2+x+1)^4 * (2x + 1)#

Therefore,
#f'(x) = g'(x)h(x) + g(x)h'(x)#
# = 8*(2x-3)^3*(x^2+x+1)^5 + (2x-3)^4 * 5(x^2+x+1)^4 * (2x+1)#
# = (2x-3)^3(8*(x^2+x+1)^5 + (2x-3) * 5(x^2+x+1)^4 * (2x+1))#
# = (2x-3)^3*(x^2+x+1)^4* (8(x^2+x+1)+5(2x-3)(2x+1))#
# = (2x-3)^3*(x^2+x+1)^4* (8x^2+8x+8+5(4x^2-x-3))#
# = (2x-3)^3*(x^2+x+1)^4* (8x^2+8x+8+20x^2-5x-15)#
# = (2x-3)^3(x^2 + x + 1)^4(28x^2-12x-7) square#

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