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

In summary: I am sorry.In summary, the conversation discusses using the definition of the derivative to find f'(x) for the function f(x)=(12)/(sqrt(1+3x)). The attempt at a solution involves using the difference quotient and the conjugate technique to simplify the expression. After some confusion and errors, the correct method is explained and the answer is obtained.
  • #1
Rayquesto
318
0

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?
 
Physics news on Phys.org
  • #2
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!
 
  • #3
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:
  • #4
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
 
  • #5
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?
 
  • #6
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.
 
  • #7
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.

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

The use the conjugate.
 
  • #8
SammyS said:
It's a difference, not a sum.

[itex]\displaystyle f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}[/itex]
[itex]\displaystyle=\lim_{h\to0}\frac{\displaystyle\frac{12}{\sqrt{1+3(x+h)}}-\frac{12}{\sqrt{1+3x}}}{h}[/itex]​
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! :)
 
  • #9
I noticed I did that earlier, but I had a simple miscalculation.
 

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

What is the definition of derivative?

The derivative of a function at a specific point is the slope of the tangent line to the graph of the function at that point.

How is the derivative of a function mathematically expressed?

The derivative of a function f(x) is mathematically expressed as f'(x) or dy/dx.

What is the relationship between the derivative and the original function?

The derivative of a function represents the rate of change of the function at a specific point, while the original function represents the values of the function at different points.

Can the derivative of a function be negative?

Yes, the derivative of a function can be negative if the slope of the tangent line is negative at a specific point, indicating a decreasing rate of change.

Why is the concept of derivative important in mathematics and science?

The derivative is important because it helps us understand the behavior of functions and their rates of change, which are essential in solving various real-world problems in fields such as physics, economics, and engineering.

Similar threads

Back
Top