anonymous
  • anonymous
Find the derivative of f(x) = 3x + 8 at x = 4.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
terenzreignz
  • terenzreignz
Interestingly (though not really)... the derivative of a linear function is just the slope.
inkyvoyd
  • 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
  • anonymous
these are the answers idk how to get it.. 1 3 4 8

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inkyvoyd
  • 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
  • terenzreignz
First things first... the derivative of a sum (or difference) is the *sum (difference) of their derivatives* Aye?
anonymous
  • anonymous
ya seriously lol i'm in pre calc..... and i think so
terenzreignz
  • terenzreignz
Okay, what's the derivative of a power? \[\Large \frac{d}{dx}ax^n=\color{red}?\]
anonymous
  • anonymous
0?
terenzreignz
  • terenzreignz
Don't guess. You seriously aren't familiar with the power rule? D:
anonymous
  • anonymous
nope i'm taking it online they don't teach anything
terenzreignz
  • 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
  • terenzreignz
And THAT... is the power rule: \[\Large \frac{d}{dx}ax^n = \color{blue}{nax^{n-1}}\]
terenzreignz
  • terenzreignz
So, having said that, what is the derivative of \[\Large \frac{d}{dx}3x = \frac{d}{dx}3x^1=\color{red}?\]
anonymous
  • anonymous
is it 1
terenzreignz
  • terenzreignz
I don't know, IS IT? Check the formula I gave you.
anonymous
  • anonymous
im so confused
inkyvoyd
  • inkyvoyd
@luv2chatxoxo , have you gone over the "difference quotient" yet?
terenzreignz
  • terenzreignz
I don't know, IS IT? Check the formula I gave you.
terenzreignz
  • 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
  • anonymous
so is it 3x^0?
terenzreignz
  • terenzreignz
Precisely ^_^ And what's x^0?
terenzreignz
  • terenzreignz
Hey, you still there? You're doing great... now what is \[\Large x^0=\color{red}? \]
anonymous
  • anonymous
1 sorry!
terenzreignz
  • 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
  • anonymous
8
anonymous
  • anonymous
wait hold on
anonymous
  • anonymous
would it be 8x^-1?
terenzreignz
  • terenzreignz
You forgot something... \[\Large \frac{d}{dx}ax^n = \color{red}nax^{n-1}\]
anonymous
  • anonymous
so would it just be 0?
terenzreignz
  • 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
  • anonymous
3!!
anonymous
  • anonymous
right? haha
terenzreignz
  • 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
  • anonymous
so we don't even have to do anything with the 4?
terenzreignz
  • 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
  • anonymous
see now would it be 8x^1 + x^-1? and do we plug in four for x?
terenzreignz
  • terenzreignz
uh-oh... how did you get x^-1? The original exponent of x was 1...
anonymous
  • anonymous
crap i mean x^0
terenzreignz
  • terenzreignz
Which is...?
anonymous
  • anonymous
1
anonymous
  • anonymous
so would the answer be 36
terenzreignz
  • terenzreignz
uhh... 8x + 1?
terenzreignz
  • terenzreignz
Maybe you need to check that again...
anonymous
  • anonymous
oh wow 33
terenzreignz
  • 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
  • anonymous
i have another problem i can send you and i can see if i get it
anonymous
  • anonymous
this one!! Find the derivative of f(x) = 8x2 + 11x at x = 7.
terenzreignz
  • terenzreignz
Well, shoot. Go ahead and do it. Tell me your answer.
anonymous
  • anonymous
123?
terenzreignz
  • terenzreignz
Brilliant ^_^
terenzreignz
  • 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
  • anonymous
ah yay thank you!! quick question though
anonymous
  • anonymous
how would you solve this one? Find the derivative of f(x) = 5/x at x = -1.
anonymous
  • anonymous
do you just plug in -1 for x?
anonymous
  • anonymous
so it would be -1^0?
terenzreignz
  • terenzreignz
Hang on. Step-by-step. What's the derivative of 5/x ? do you know?
anonymous
  • anonymous
noo
terenzreignz
  • 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
  • anonymous
how do you get to the negative 1 though
anonymous
  • anonymous
like x raised to it
terenzreignz
  • terenzreignz
Seriously? LOL laws of exponents... \[\Large \frac1{x^n}= x^{-n}\]
anonymous
  • anonymous
haah ok ok i remember now ive had a long morning okay so then i plug in the -1 for x correct?
terenzreignz
  • terenzreignz
Not yet. First find the derivative.
anonymous
  • anonymous
why am i getting so confsued?
terenzreignz
  • terenzreignz
Because... you didn't pay attention to basics? :P
anonymous
  • anonymous
nooo i just have to teach everything to myself lol so 5x^-1 idk what to do
terenzreignz
  • terenzreignz
I said... THIS \[\Large \frac{d}{dx}ax^n = \color{blue}{nax^{n-1}}\] still applies! Even if the exponent is negative!
anonymous
  • anonymous
ohh!!!! so 5x^-2
terenzreignz
  • terenzreignz
No... remember that your n is negative! Pay attention to signs!
anonymous
  • anonymous
so wouldnt that be -1-1? isn't that negative 2?
terenzreignz
  • terenzreignz
You can't keep forgetting these things! \[\Large \frac{d}{dx}ax^n = \color{red}nax^{n-1}\]
anonymous
  • anonymous
-5x^-2?
terenzreignz
  • terenzreignz
Better. Now plug in x = -1
anonymous
  • anonymous
so -5?
terenzreignz
  • terenzreignz
mhmm ^_^
anonymous
  • 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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