Question

Find the derivative of f(x) = 4 divided by x at x = 2.

Answer 1

ANSWER

$f'(2) = -1$

EXPLANATION

The given function is

$f(x) = \frac{4}{x}$

Recall that:

$\frac{c}{ {a}^{ m} } = c {a}^{ - m}$

We rewrite the given function using this rule to obtain,

$f(x) = 4 {x}^{ - 1}$

Recall again that,

If

$f(x)= a {x}^{n}$

then

$f'(x)=n a {x}^{n - 1}$

We differentiate using the power rule to obtain,

$f'(x) = - 1 \times 4 {x}^{ - 1 - 1}$

$f'(x) = - 4 {x}^{ - 2}$

We rewrite as positive index to obtain,

$f'(x) = - \frac{4}{ {x}^{2} }$

We plug in x=2 to obtain,

$f'(2) = - \frac{4}{ { (2)}^{2} } = - \frac{4}{4} = - 1$

Answer 2

Hello!

The answer is:

$f'(2)=-1$

Why?

To solve this problem, first we need to derivate the given function, and then, evaluate the derivated function with x equal to 2.

The given function is:

$f(x)=\frac{4}{x}$

It's a quotient, so, we need to use the following formula to derivate it:

$f'(x)=\frac{d}{dx}(\frac{u}{v}) =\frac{v*u'-u*v'}{v^{2} }$

Then, of the given function we have that:

$u=4\\v=x$

So, derivating we have:

$f'(x)=\frac{d}{dx}(\frac{4}{x}) =\frac{x*(4)'-4*(x)'}{x^{2} }$

$f'(x)=\frac{d}{dx}(\frac{4}{x}) =\frac{x*0-4*1}{x^{2} }$

$f'(x)=\frac{d}{dx}(\frac{4}{x}) =\frac{0-4}{x^{2} }$

$f'(x)=\frac{d}{dx}(\frac{4}{x}) =\frac{-4}{x^{2} }$

Hence,

$f'(x)=\frac{d}{dx}(\frac{4}{x}) =\frac{-4}{x^{2} }$

Now, evaluating with x equal to 2, we have:

$f'(2)=\frac{-4}{(2)^{2} }$

$f'(2)=\frac{-4}{4}$

$f'(2)=-1$

Therefore, the answer is:

$f'(2)=-1$

Have a nice day!