What is the Definition of the Derivative for f(x)=(12)/(sqrt(1+3x))?

Rayquesto
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Homework Statement



Use the definition of the derivative to find f'(x) is f(x)=(12)/(sqrt(1+3x))

Homework Equations



I have a feeling you use some conjugate techniques, but I'm stuck.

f'(x)=lim h->0 (f(x+h) - f(x))/(h)

The Attempt at a Solution



I could post a very long list of useless equations, but that would be wasting time. I don't know how to do this or where to start, since I know what I'm doing is not the correct way to start. I know what the answer is using the quotient and power rules, but they want you to use the definition of the derivative. So, the start would be to state that lim h->0 (f(x+h) - f(x))/(h), but there's algebra involved that I'm missing. Anyone want to help?
 
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Well, for one thing, you might want to say what f(x) is. Now write out the difference quotient and yes, multiply by the conjugate. At least show where you are starting from!
 
Thank you so much for the reply. So, the difference quotient for this function is:

(12/(sqrt(1+3x+3h))-(12/sqrt(1+3x))/(h)

so,

f'(x)=lim h->0 (12/hsqrt(1+3x+3h)) - (12/hsqrt(1+3)) *OK OK OK It's minus! Simple error.

How do I utilize the conjugate technique? I thought I knew how to use it, but I ended up with something wrong. I tried a myriad of ways I thought you would use the conjugate technique. I even tried multiplying by "1" such that hsqrt(1+3x+3h)/hsqrt(1+3x+3h)=1 and hsqrt(1+3x)/hsqrt(1+3x)=1 and used the sum limit law and used the conjugate and I even combined the two equations using some algebra. So, where do I begin to use the conjugate technique? Quite possibly, I'm missing the true idea of the technique of conjugate. I see it as taking not necessarily the negative or 1 and multiplying it by the function, but taking what could be a and -bi and multiplying that as an equation such that you use the numerator and assign a and -bi and multiplying by 1.
 
Last edited:
Rayquesto said:
Thank you so much for the reply. So, the difference quotient for this function is:

(12/(sqrt(1+3x+3h))+(12/sqrt(1+3x))/(h)


Is it difference?

ehild
 
Oh thanks for catching that. It's what happens when I got a lot on my mind while I type. The same goes with spelling errors, but I'm working on the simple errors. I will eventually get it all down, however, pointing that error out really doesn't help get me anywhere with trying to understand ideas that I don't really understand. That right there was clearly one of those simple errors that didn't influence my answer. I wrote the "-" sign on my paper and was consistent throughout my calculations. So, it did not influence my true errors that I need to fix for the problem itself. So, my question should be where do I start with the conjugate technique?
 
so, here's the correction:f'(x)=lim h->0 (12/hsqrt(1+3x+3h)) - (12/hsqrt(1+3))

As you could tell from the beginning of the question, I wrote down the formula with the "-" sign. So, that clearly was a simple error that did not influence my errors in conjugate endeavors.
 
Rayquesto said:
Thank you so much for the reply. So, the difference quotient for this function is:

(12/(sqrt(1+3x+3h))+(12/sqrt(1+3x))/(h)

so,

f'(x)=lim h->0 (12/hsqrt(1+3x+3h)) + (12/hsqrt(1+3))

How do I utilize the conjugate technique? I thought I knew how to use it, but I ended up with something wrong. I tried a myriad of ways I thought you would use the conjugate technique. I even tried multiplying by "1" such that hsqrt(1+3x+3h)/hsqrt(1+3x+3h)=1 and hsqrt(1+3x)/hsqrt(1+3x)=1 and used the sum limit law and used the conjugate and I even combined the two equations using some algebra. So, where do I begin to use the conjugate technique? Quite possibly, I'm missing the true idea of the technique of conjugate. I see it as taking not necessarily the negative or 1 and multiplying it by the function, but taking what could be a and -bi and multiplying that as an equation such that you use the numerator and assign a and -bi and multiplying by 1.
It's a difference, not a sum.

\displaystyle f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}
\displaystyle=\lim_{h\to0}\frac{\displaystyle\frac{12}{\sqrt{1+3(x+h)}}-\frac{12}{\sqrt{1+3x}}}{h}​
Use a common denominator to combine the fractions in the numerator.

The use the conjugate.
 
SammyS said:
It's a difference, not a sum.

\displaystyle f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}
\displaystyle=\lim_{h\to0}\frac{\displaystyle\frac{12}{\sqrt{1+3(x+h)}}-\frac{12}{\sqrt{1+3x}}}{h}​
Use a common denominator to combine the fractions in the numerator.

The use the conjugate.

Thank you! I realize the mall error I had, but please read the comments from earlier. We covered that earlier. And ok yes you are the one! Thank you for telling me how it's done. I got it now! :)
 
I noticed I did that earlier, but I had a simple miscalculation.
 
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