Find the derivative of f(x) = 3x + 8 at x = 4.

- anonymous

Find the derivative of f(x) = 3x + 8 at x = 4.

- Stacey Warren - Expert brainly.com

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- schrodinger

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- terenzreignz

Interestingly (though not really)... the derivative of a linear function is just the slope.

- inkyvoyd

It's delta y/delta x as delta y and delta x get close to zero - but for a linear equation, the slope is always constant - aka the rate of change

- anonymous

these are the answers idk how to get it..
1
3
4
8

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## More answers

- inkyvoyd

word of unrelated advice - learn calculus. Seriously, if you're in AB or BC and you decide for whatever reason not to pay attention during limits and derivatives, you're going to fail integrals completely

- terenzreignz

- anonymous

ya seriously lol i'm in pre calc..... and i think so

- terenzreignz

Okay, what's the derivative of a power?
\[\Large \frac{d}{dx}ax^n=\color{red}?\]

- anonymous

0?

- terenzreignz

Don't guess.
You seriously aren't familiar with the power rule? D:

- anonymous

nope i'm taking it online they don't teach anything

- terenzreignz

Here it is...
\[\Large \frac{d}{dx}ax^n\]
You take the exponent (n), multiply it to the function:
\[\Large \frac{d}{dx}ax^n\rightarrow \color{red}nax^n\]
BUT you subtract 1 from the exponent
\[\Large \frac{d}{dx}ax^n\rightarrow \color{red}nax^n\rightarrow nax^{\color{red}{n-1}}\]

- terenzreignz

And THAT... is the power rule:
\[\Large \frac{d}{dx}ax^n = \color{blue}{nax^{n-1}}\]

- terenzreignz

So, having said that, what is the derivative of
\[\Large \frac{d}{dx}3x = \frac{d}{dx}3x^1=\color{red}?\]

- anonymous

is it 1

- terenzreignz

I don't know, IS IT?
Check the formula I gave you.

- anonymous

im so confused

- inkyvoyd

@luv2chatxoxo , have you gone over the "difference quotient" yet?

- terenzreignz

I don't know, IS IT?
Check the formula I gave you.

- terenzreignz

Okay, here's an example:
Suppose I want the derivative of \(\large 5x^4\).
So first, the exponent is 4, so I multiply that to the function, but I subtract 1 from the exponent... like so:
\[\Large \frac{d}{dx}5x^4 = \color{red}4\cdot5x^{4\color{red}{-1}}=\color{blue}{20x^3}\]

- anonymous

so is it 3x^0?

- terenzreignz

Precisely ^_^
And what's x^0?

- terenzreignz

Hey, you still there?
You're doing great...
now what is
\[\Large x^0=\color{red}? \]

- anonymous

1 sorry!

- terenzreignz

Good.
So the derivative of 3x is just 3.
What about the derivative of 8?
Clue:
\[\Large 8 = 8x^0\]
Apply the power rule again...

- anonymous

8

- anonymous

wait hold on

- anonymous

would it be 8x^-1?

- terenzreignz

You forgot something...
\[\Large \frac{d}{dx}ax^n = \color{red}nax^{n-1}\]

- anonymous

so would it just be 0?

- terenzreignz

That's precisely it :P
Keep that in mind.
The derivative of a *constant* is *always* zero.
So the derivative of 3x + 8
is just the sum of the derivatives of 3x and 8
Which are 3 and 0, respectively.
so finally, what's the derivative of 3x + 8?

- anonymous

3!!

- anonymous

right? haha

- terenzreignz

Yes.
And 3 is constant, so no matter what x is, be it 4, or literally anything on the domain, the derivative is always 3.

- anonymous

so we don't even have to do anything with the 4?

- terenzreignz

Nope.
You got lucky, the derivative of this one was a constant.
Just so I know you understand, what's the derivative of
\[\Large 4x^2+x\]at x = 4?

- anonymous

see now would it be 8x^1 + x^-1? and do we plug in four for x?

- terenzreignz

uh-oh...
how did you get x^-1?
The original exponent of x was 1...

- anonymous

crap i mean x^0

- terenzreignz

Which is...?

- anonymous

1

- anonymous

so would the answer be 36

- terenzreignz

uhh...
8x + 1?

- terenzreignz

Maybe you need to check that again...

- anonymous

oh wow 33

- terenzreignz

tsk tsk :P
One last, just to reassure me (and yourself) that you're well on your way to mastering the power rule.

- anonymous

i have another problem i can send you and i can see if i get it

- anonymous

this one!! Find the derivative of f(x) = 8x2 + 11x at x = 7.

- terenzreignz

Well, shoot.
Go ahead and do it.
Tell me your answer.

- anonymous

123?

- terenzreignz

Brilliant ^_^

- terenzreignz

Try this one too.
I have confidence in you...
Derivative of \[\Large x^3 +2x^2-6x+4\] at x = 1
Go for it, tiger :P

- anonymous

ah yay thank you!! quick question though

- anonymous

how would you solve this one? Find the derivative of f(x) = 5/x at x = -1.

- anonymous

do you just plug in -1 for x?

- anonymous

so it would be -1^0?

- terenzreignz

Hang on.
Step-by-step.
What's the derivative of 5/x ?
do you know?

- anonymous

noo

- terenzreignz

Pity :P
Let's write it this way:
\[\Large \frac5x = 5x^{-1}\]
does that look better? :P
Power rule still applies, by the way..

- anonymous

how do you get to the negative 1 though

- anonymous

like x raised to it

- terenzreignz

Seriously? LOL
laws of exponents...
\[\Large \frac1{x^n}= x^{-n}\]

- anonymous

haah ok ok i remember now ive had a long morning okay so then i plug in the -1 for x correct?

- terenzreignz

Not yet.
First find the derivative.

- anonymous

why am i getting so confsued?

- terenzreignz

Because... you didn't pay attention to basics? :P

- anonymous

nooo i just have to teach everything to myself lol so 5x^-1 idk what to do

- terenzreignz

I said... THIS
\[\Large \frac{d}{dx}ax^n = \color{blue}{nax^{n-1}}\]
still applies!
Even if the exponent is negative!

- anonymous

ohh!!!! so 5x^-2

- terenzreignz

No... remember that your n is negative!
Pay attention to signs!

- anonymous

so wouldnt that be -1-1? isn't that negative 2?

- terenzreignz

You can't keep forgetting these things!
\[\Large \frac{d}{dx}ax^n = \color{red}nax^{n-1}\]

- anonymous

-5x^-2?

- terenzreignz

Better.
Now plug in x = -1

- anonymous

so -5?

- terenzreignz

mhmm ^_^

- anonymous

one more :( i'm sorry i jsut need to make sure i get these Find the derivative of f(x) = -7/x at x = -3.

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