How do you differentiate #f(x)=x^2(x+7)^3# using the product rule?

2 Answers
Nov 7, 2016

Answer:

#f'(x)=(x+7)^2(x)(5x+14)#

Explanation:

#f(x)=(x)^2(x+7)^3#

Product rule: if #f(x) = AB#, then #f'(x)=AB' + A'B#

In this case, too differentiate terms "A" and "B", we have to use the chain rule.

#f'(x)=(x)^2(3)(x+7)^2+(2x)(x+7)^3#

Now simplify by factoring:
#f'(x)=(x+7)^2(x)[3x+2(x+7)]#
#f'(x)=(x+7)^2(x)(3x+2x+14)#
#f'(x)=(x+7)^2(x)(5x+14)#

Nov 7, 2016

Answer:

# :. f'(x) = x(x+7)^2 (5x+14} #

Explanation:

If you are studying maths, then you should learn the Product Rule for Differentiation, and practice how to use it:

# d/dx(uv)=u(dv)/dx+v(du)/dx #, or, # (uv)' = (du)v + u(dv) #

I was taught to remember the rule in words; "The first times the derivative of the second plus the second times the derivative of the first ".

So with # f(x)=x^2(x+7)^3 # Then

# { ("Let "u=x^2, => , (du)/dx=2x), ("And "v=(x+7)^3, =>, (dv)/dx=3(x+7)^2 " (by chain rule)" ) :}#

# :. d/dx(uv) = u(dv)/dx+v(du)/dx #
# :. f'(x) = (x^2)(3(x+7)^2) + ((x+7)^3)(2x) #
# :. f'(x) = 3x^2(x+7)^2 + 2x(x+7)^3 #
# :. f'(x) = x(x+7)^2 {3x+2(x+7)} #
# :. f'(x) = x(x+7)^2 (3x+2x+14} #
# :. f'(x) = x(x+7)^2 (5x+14} #