# How do you use the Product Rule to find the derivative of (x^2)(x^3+4)?

The explanation is given below.

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for differentiation where product of functions is given.

If ##f(x)## and ##g(x)## are two functions then their product would be ##f(x)*g(x)##

The derivative ##(f(x)*g(x))’## is given by the product rule

##(f(x)g(x))’ = f(x)g'(x)+g(x)f'(x)##

Derivative rules

##d/dx (x^n) = nx^(n-1)##
##d/dx(f(x)+g(x)) = d/dx(f(x)) + d/dx(g(x))##
##d/dx(c) = 0## Derivative of constant is zero.

Now coming to our problem

##y=(x^2)(x^3+4)##

##f(x) = x^2## and ##g(x) = (x^3+4)##

##f'(x) = d/dx(x^2) => f'(x) = 2x##
##g'(x) = d/dx(x^3+4) = d/dx (x^3) + d/dx(4)##
##g'(x) = 3x^2 + 0##
##g'(x) = 3x^2##

Using product rule

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

##= x^2(3x^2) + (x^3+4)(2x)##

##= 3x^4 + 2x^4+8x## simplifying

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