Find the slope of the function f(x)= sqrt(5x) at x=5

Question

anonymous · Answer

$$f(x) = \sqrt{5x}$$

anonymous · Answer

You have to find the derivative of f(x). Then you have to calculate the value of the derivative for x=5, that is the f'(5).

anonymous · Answer

How do I do that? Like how do I find the derivative?

anonymous · Answer

Would I substitute in 5? That would give me f(x) = 5

anonymous · Answer

Have you learned derivatives?

anonymous · Answer

Yes I have. but I dont know what I'm doing, to be honest. I'm so lost in class. But Wouldn't you have to put sqrt5x into 5? thats what your supposed to do, right? but you can't. thats why I'm confused

anonymous · Answer

To find the slope, you have to find the derivative of the function first. The derivative is denoted by f'(x). For

$$f(x)  = \sqrt{5x} = \sqrt{5}\sqrt{x} $$
you have:
$$f'(x) = (f(x))' = (\sqrt{5}\sqrt{x})' = \sqrt{5}(\sqrt{x})' = \sqrt{5} \frac{1}{2\sqrt{x}} = \frac{\sqrt{5}}{2\sqrt{x}}$$

Then to find the slope at x=5, you have to find f'(5), that is, in the derivative formula you
replace x with 5.

anonymous · Answer

so would I end up with 25/2

anonymous · Answer

$$f'(5) = \frac{\sqrt{5}}{2\sqrt{5}} = ..$$

anonymous · Answer

whoops, that would make it 1/2

phi · Answer

are you doing multiple guess ?

you could write
$$ \sqrt{5x} =  \sqrt{5} x^{\frac{1}{2}} $$
and use the rule
$$ \frac{ d x^n}{dx} = n x^{n-1} $$

anonymous · Answer

That's right @cutie.patootie !

anonymous · Answer

Oh sweet! Thanks!!!

anonymous · Answer

that was so confusing.. Lol

anonymous · Answer

Do you know what should be done if you want to find the slope at x=20 now?

anonymous · Answer

I would just replace x with 20, right?

anonymous · Answer

Yes, you would replace x with 20 in  the f'(x) formula. So what would you get?

anonymous · Answer

1/10

anonymous · Answer

$$f'(x) = \frac{\sqrt{5}}{2\sqrt{x}} $$
Replacing x with 20:
$$f'(20) = \frac{\sqrt{5}}{2\sqrt{20}} $$
$$f'(20) = \frac{\sqrt{5}}{2\sqrt{4*5}} $$
$$f'(20) = \frac{\sqrt{5}}{2\sqrt{4}*\sqrt{5}} $$
$$f'(20) = \frac{\sqrt{5}}{2*2\sqrt{5}} $$
$$f'(20) = \frac{\sqrt{5}}{4\sqrt{5}} $$
$$f'(20) = \frac{1}{4}$$

anonymous · Answer

Ohhh. I didn't separate the 20

anonymous · Answer

It's ok, you will get used to it. In general you want to siplify the square root, so you think how you can separate the value inside the square root, so you can pull something out of the root. The square root of 4 is 2, so you could write 20=4*5.

anonymous · Answer

okay, thank you for your help:))))

anonymous · Answer

Anytime!