Help with Derivative Homework: f(x)=2x+4/√x

[Krypto78]

Homework Statement

Given the equation f(x) = 2x+4/√x evaluate:

(1) f(0.5)
(2) f '(0.5)

Homework Equations

The Attempt at a Solution

(1) is simply to substitute the value 0.5 into the equation

(2) however i am assuming is to find the derivative of the equation

This is where i get stuck. Its so basic i know but i always seem to struggle with the simple things.

this is where I am at

f(x) = 2x+4/√x
= 2x/√x + 4/√x

The reciprocal of√x is x^1/2 so I am assuming the 2x/√x is 2x(x)^-0.5 and this is how far i get on this part of the equation.

The 4/√x I am sure = 4x^-0.5 and using the 'direct rule' becomes:-

(-0.5)(4)x^-0.5 - 1

converting to root from:-

-2x^-2/3

from there i get lost.

I would greatly appreciate a step by step solution.

Thanks guys

 

[jbunniii]

You wrote f(x) = 2x + 4/sqrt(x), which would equal

$$f(x) = 2x + \frac{4}{\sqrt{x}}$$

under the standard rules of operator precedence. However, based on what you subsequently wrote, I am guessing the function is actually

$$f(x) = \frac{2x + 4}{\sqrt{x}}$$

which, if you wanted to write it "in line", would be f(x) = (2x + 4)/sqrt(x) [note the parentheses]. This can be simplified as

$$f(x) = \frac{2x}{\sqrt{x}} + \frac{4}{\sqrt{x}} = 2\sqrt{x} + \frac{4}{\sqrt{x}}$$

Now, what is this in terms of powers of x?

 

[Krypto78]

ill have a go

2x^0.5 + 4x^-0.5

am i on the right path?

 

[jbunniii]

ill have a go

2x^0.5 + 4x^-0.5

am i on the right path?

Yes, that's right. So what do you get if you differentiate this?

 

[Mark44]

Homework Statement

Given the equation f(x) = 2x+4/√x evaluate:

As others have noted already, the right side needs to be written as (2x + 4)/√x.

(1) f(0.5)
(2) f '(0.5)

Homework Equations

The Attempt at a Solution

(1) is simply to substitute the value 0.5 into the equation

(2) however i am assuming is to find the derivative of the equation

This is where i get stuck. Its so basic i know but i always seem to struggle with the simple things.

this is where I am at

f(x) = 2x+4/√x
= 2x/√x + 4/√x

The reciprocal of√x is x^1/2 so I am assuming the 2x/√x is 2x(x)^-0.5 and this is how far i get on this part of the equation.

No, the reciprocal of √x is NOT "x^1/2". The reciprocal of √x is 1/√x, or x^(-1/2). √x and x^(1/2) - note the parentheses) are two ways of writing the same thing.

The 4/√x I am sure = 4x^-0.5 and using the 'direct rule' becomes:-

(-0.5)(4)x^-0.5 - 1

Please use parentheses around your exponents, like so:
(-0.5)(4)x^(-0.5 - 1)

converting to root from:-

-2x^-2/3

and here, too.

-2x^(-2/3)

from there i get lost.

I would greatly appreciate a step by step solution.

That's not the way it works here at Physics Forums. The rules don't permit us to provide you with a step-by-step solution, but we're happy to guide you as you do the work.

 

[Krypto78]

Thanks for the reply guys.

Totally understand Mark, thanks for your input:)

jbunniii can you go back 1 step. You simplified 2x/√x to be 2√x. Can i ask how you got this?

on a side note i am doing a self study course so i am my own tutor at present. I really want to get a handle on this so please bare with me.

Thanks

 

[HallsofIvy]

Thanks for the reply guys.

Totally understand Mark, thanks for your input:)

jbunniii can you go back 1 step. You simplified 2x/√x to be 2√x. Can i ask how you got this?

If jbunniii will forgive me for jumping in here there are two ways to do that:
first, note that $x= (\sqrt{x})^2= (\sqrt{x})(\sqrt{x})$ so that
$$\frac{x}{\sqrt{x}}= \frac{(\sqrt{x})(\sqrt{x})}{\sqrt{x}}= \sqrt{x}$$
or
$$x/\sqrt{x}= x^1/x^{1/2}= x^1x^{-1/2}= x^{1- 1/2}= x^{1/2}= \sqrt{x}$$

on a side note i am doing a self study course so i am my own tutor at present. I really want to get a handle on this so please bare with me.

Thanks

Oh, dear. Another person who wants us to "bare" with him. This forum is getting too wild for me!

 

[Krypto78]

Oh boy..."bare".

Is it too late to change that to "bear" :)

 

[Krypto78]

Right where were we

Yes, that's right. So what do you get if you differentiate this?

2x^(0.5) + 4x^(-0.5)

= (1/2)(2)x^(0.5 -1) + (-1/2)(4)x^(-1/2 - 1)

= x^(-1/2) + -2x^(-2/3)

Am i right so far

 

[jbunniii]

Right where were we



2x^(0.5) + 4x^(-0.5)

= (1/2)(2)x^(0.5 -1) + (-1/2)(4)x^(-1/2 - 1)

= x^(-1/2) + -2x^(-2/3)

Am i right so far

Almost. What is -1/2 - 1? (Hint: it's not -2/3.)

 

[Krypto78]

Almost. What is -1/2 - 1? (Hint: it's not -2/3.)

Hmmmmm now i am stumped.I think this is where I am going wrong. Going by the calculator it is -2/3. Is it 2/3( -and- = + thus 1/2 + 1)

i feel like a plonker:(

 

[Mark44]

Hmmmmm now i am stumped.I think this is where I am going wrong. Going by the calculator it is -2/3. Is it 2/3( -and- = + thus 1/2 + 1)

i feel like a plonker:(

Better get another calculator...

If you have 1/2 pizza and 1 pizza, that's not the same as 2/3 pizza.

 

[Krypto78]

Better get another calculator...

If you have 1/2 pizza and 1 pizza, that's not the same as 2/3 pizza.

wow it really has been a long day.

ok ok. Round the wrong way.

so its -3/2

 

[Krypto78]

Almost. What is -1/2 - 1? (Hint: it's not -2/3.)

with the correction:-

x^(-1/2) + -2x^(-3/2)

x^(-1/2) = 1/√x

2x^(-3/2) this is where I am lost.....

Thanks for the guidance guys

 

[jbunniii]

with the correction:-

x^(-1/2) + -2x^(-3/2)

x^(-1/2) = 1/√x

2x^(-3/2) this is where I am lost.....

Thanks for the guidance guys

Well, x^(-1/2) - 2x^(-3/2) is a perfectly valid answer. If you don't like negative roots, then use the fact that

$$x^{-t} = \frac{1}{x^t}$$

[edit] I just re-read the original post and recalled that it asks for f'(0.5). So you need to plug in x = 0.5, and simplify if desired.

 

[Krypto78]

Well, x^(-1/2) - 2x^(-3/2) is a perfectly valid answer. If you don't like negative roots, then use the fact that

$$x^{-t} = \frac{1}{x^t}$$

[edit] I just re-read the original post and recalled that it asks for f'(0.5). So you need to plug in x = 0.5, and simplify if desired.

So just to finish off, removing the negative roots would =

1/√x - 2/x^(3/2) ?

 

[jbunniii]

So just to finish off, removing the negative roots would =

1/√x - 2/x^(3/2) ?

Looks right to me. Now you need to evaluate this at x = 1/2.

 

[Krypto78]

Thank you SO much for your help.

What a great forum :)

 

[Krypto78]

Well, x^(-1/2) - 2x^(-3/2) is a perfectly valid answer. If you don't like negative roots, then use the fact that

$$x^{-t} = \frac{1}{x^t}$$

[edit] I just re-read the original post and recalled that it asks for f'(0.5). So you need to plug in x = 0.5, and simplify if desired.

Before we close this chapter, can i ask what is the "name" given when removing the negative roots?

 

[Mark44]

Far as I know, there's no name. What jbunniii showed is a property of exponents.