Question

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

Answer 1

`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)`

Answer 2

`:. 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}`