Respuesta :
Hello!
The answer is:
[tex]f'(2)=-1[/tex]
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:
[tex]f(x)=\frac{4}{x}[/tex]
It's a quotient, so, we need to use the following formula to derivate it:
[tex]f'(x)=\frac{d}{dx}(\frac{u}{v}) =\frac{v*u'-u*v'}{v^{2} }[/tex]
Then, of the given function we have that:
[tex]u=4\\v=x[/tex]
So, derivating we have:
[tex]f'(x)=\frac{d}{dx}(\frac{4}{x}) =\frac{x*(4)'-4*(x)'}{x^{2} }[/tex]
[tex]f'(x)=\frac{d}{dx}(\frac{4}{x}) =\frac{x*0-4*1}{x^{2} }[/tex]
[tex]f'(x)=\frac{d}{dx}(\frac{4}{x}) =\frac{0-4}{x^{2} }[/tex]
[tex]f'(x)=\frac{d}{dx}(\frac{4}{x}) =\frac{-4}{x^{2} }[/tex]
Hence,
[tex]f'(x)=\frac{d}{dx}(\frac{4}{x}) =\frac{-4}{x^{2} }[/tex]
Now, evaluating with x equal to 2, we have:
[tex]f'(2)=\frac{-4}{(2)^{2} }[/tex]
[tex]f'(2)=\frac{-4}{4}[/tex]
[tex]f'(2)=-1[/tex]
Therefore, the answer is:
[tex]f'(2)=-1[/tex]
Have a nice day!
ANSWER
[tex]f'(2) = -1[/tex]
EXPLANATION
The given function is
[tex]f(x) = \frac{4}{x} [/tex]
Recall that:
[tex] \frac{c}{ {a}^{ m} } = c {a}^{ - m} [/tex]
We rewrite the given function using this rule to obtain,
[tex]f(x) = 4 {x}^{ - 1} [/tex]
Recall again that,
If
[tex]f(x)= a {x}^{n} [/tex]
then
[tex]f'(x)=n a {x}^{n - 1} [/tex]
We differentiate using the power rule to obtain,
[tex]f'(x) = - 1 \times 4 {x}^{ - 1 - 1} [/tex]
[tex]f'(x) = - 4 {x}^{ - 2} [/tex]
We rewrite as positive index to obtain,
[tex]f'(x) = - \frac{4}{ {x}^{2} } [/tex]
We plug in x=2 to obtain,
[tex]f'(2) = - \frac{4}{ { (2)}^{2} } = - \frac{4}{4} = - 1[/tex]