Use the quotient rule to evaluate h′(a) for the given function h(x) and a. h(x)=4x/x+4 a=−3

Question

please reply with answer.

Answer 1

Let's break up the problem.

When is a quotient rule needed to be applied? When we have a numerator/denominator or 2 quantities/functions divided. Hence our case h(x) = f(x)/g(x) where f(x) = 4x and g(x) = x+4, hence

h(x) = 4x/(x+4)

What is a quotient rule? Answer:

h'(x) = [f'(x)g(x)-g'(x)f(x)]/(g(x))²

Note here how everything looks jumbled, but let's break up each component, we already know f(x) and g(x) now let's take the derivative separately f'(x)=4 and g'(x) = 1 (remember derivative of a constant is 0 and derivative of x = 1)

Now let's substitute in our quotient rule from before:

h'(x) = [4(x+4)-(4x)]/(x+4)²

Simplify by first factoring the numerator: 4x+16-4x = 16, therefore we are left with:

h'(x)=16/(x+4)²

Now the question asks for a given value of x, i.e. when x = -3, but the wording is h'(a) when a is given, a=-3.

Let's plug 3 in for a (or I think of it as x)

h'(-3)=16/(-3+4)²

h'(-3)=16 (since the denominator before ends up being 1²)