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

1 Answer
Dec 25, 2015

Answer:

The explanation is given below.

Explanation:

Product rule 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

#=5x^4+8x# Answer