Must use definition of derivative to solve

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simpleitkeeps

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So I have been given 5 problems similar to the example I am going to provide that I must find f'(x). The only catch is I cant use any short cuts and must use the definition of the derivative to solve. I am having an extremely hard time with this. Thanks in advance for any help provided:

Find f'(x) using the definition of the derivative: f(x)=(3)/(x^5)
 
Hello, and welcome to FMH! :)

Let's begin with the definition of the derivative:

[MATH]f'(x)\equiv\lim_{h\to0}\frac{f(x+h)-f(x)}{h}[/MATH]
In this case, we're given:

[MATH]f(x)=\frac{3}{x^5}[/MATH]
And so:

[MATH]f'(x)=\lim_{h\to0}\frac{\dfrac{3}{(x+h)^5}-\dfrac{3}{x^5}}{h}[/MATH]
The first thing I would do is factor the constant out in front of the limit:

[MATH]f'(x)=3\lim_{h\to0}\frac{\dfrac{1}{(x+h)^5}-\dfrac{1}{x^5}}{h}[/MATH]
Now, in the numerator of the difference quotient, I would try to combine the two terms. What do you get when you do so?
 
Can you please show us what you have done so far? We need to work with you step by step.

I feel that you need to be able to say in words what the function f does to what is in the parenthesis. The answer is it raises what you put inside the parenthesis to the fifth power and puts that result in the denominator. It then puts 3 in the numerator.
So what is f(x+h) equal to?

By the way the whole goal is to cancel out h/h. After doing that you simply put 0 in for h.
 
I have no idea how to type as you have on here, I apologize...but I've taken a picture of what I've been attemptingIMG_20190318_223107.jpg
 
Where did the last line come from?
The 2nd to the last line on the left is correct, so please continue.
Also showing your work this way is the best way ( at least in my opinion)

Just out of curiosity where did (a+b)^5 come from??
 
Jomo said:
Where did the last line come from?
The 2nd to the last line on the left is correct, so please continue.
Also showing your work this way is the best way ( at least in my opinion)

Just out of curiosity where did (a+b)^5 come from??
Click to expand...
haha sorry the last line was me distributing it all out, it really has nothing to do with it. It is just a way to get my brain working. Disregard.
 
Let's look only at the numerator of the difference quotient I wrote:

[MATH]\frac{1}{(x+h)^5}-\frac{1}{x^5}[/MATH]
Combining terms, we get:

[MATH]\frac{x^5-(x+h)^5}{(x(x+h))^5}[/MATH]
Using the binomial theorem, this becomes:

[MATH]\frac{x^5-\left(x^5+5x^4h+10x^3h^2+10x^2h^3+5xh^4+h^5\right)}{(x(x+h))^5}[/MATH]
Now, distribute the negative sign, combine like terms and what does the entire difference quotient look like?
 
Jomo said:
Where did the last line come from?
The 2nd to the last line on the left is correct, so please continue.
Also showing your work this way is the best way ( at least in my opinion)

Just out of curiosity where did (a+b)^5 come from??
Click to expand...
(a+b)^5?? I didnt even realize.. I was doing
Jomo said:
Where did the last line come from?
The 2nd to the last line on the left is correct, so please continue.
Also showing your work this way is the best way ( at least in my opinion)

Just out of curiosity where did (a+b)^5 come from??
Click to expand...
this was what I was doing before,I have no idea why I wrote the (a+b) my brain just threw it in there
15529634372467962695921504866695.jpg
 
MarkFL said:
Let's look only at the numerator of the difference quotient I wrote:

[MATH]\frac{1}{(x+h)^5}-\frac{1}{x^5}[/MATH]
Combining terms, we get:

[MATH]\frac{x^5-(x+h)^5}{(x(x+h))^5}[/MATH]
Using the binomial theorem, this becomes:

[MATH]\frac{x^5-\left(x^5+5x^4h+10x^3h^2+10x^2h^3+5xh^4+h^5\right)}{(x(x+h))^5}[/MATH]
Now, distribute the negative sign, combine like terms and what does the entire difference quotient look like?
Click to expand...
see thats the direction i was heading, thats why the last line was distributed.. but i didn't factor out the 3 so it was screwing me up. Give me a second to work this out please. Thank you
 
It is not funny because you do have to distribute the numerator out. If you multiply (x+h) five times you should get at most x^5
 
simpleitkeeps said:
see thats the direction i was heading, thats why the last line was distributed.. but i didn't factor out the 3 so it was screwing me up. Give me a second to work this out please. Thank you
Click to expand...

It's not necessary to factor out the 3, but it does make the numbers you're working with smaller. :)
 
It's not necessary to expand the binomial in the denominator...I wind up with (I factored the -1 out front too):

[MATH]f'(x)=-3\lim\frac{5x^4h+10x^3h^2+10x^2h^3+5xh^4+h^5}{h(x(x+h))^5}[/MATH]
Next, factor the numerator:

[MATH]f'(x)=-3\lim\frac{h\left(5x^4+10x^3h+10x^2h^2+5xh^3+h^4\right)}{h(x(x+h))^5}[/MATH]
Divide out the common factor:

[MATH]f'(x)=-3\lim\frac{5x^4+10x^3h+10x^2h^2+5xh^3+h^4}{(x(x+h))^5}[/MATH]
Now, you can substitute \(h=0\) to get the value of the limit...
 
Please note that h/h is NOT 1. h/h is actually a piecewise function. In fact h/h = 1 if h is not 0
And h/h is indetermined if h=0
Since we have lim as h goes to 0, we never have h=0. So h/h = 1 or you can say it cancels out.
 
MarkFL said:
It's not necessary to expand the binomial in the denominator...I wind up with (I factored the -1 out front too):

[MATH]f'(x)=-3\lim\frac{5x^4h+10x^3h^2+10x^2h^3+5xh^4+h^5}{h(x(x+h))^5}[/MATH]
Next, factor the numerator:

[MATH]f'(x)=-3\lim\frac{h\left(5x^4+10x^3h+10x^2h^2+5xh^3+h^4\right)}{h(x(x+h))^5}[/MATH]
Divide out the common factor:

[MATH]f'(x)=-3\lim\frac{5x^4+10x^3h+10x^2h^2+5xh^3+h^4}{(x(x+h))^5}[/MATH]
Now, you can substitute \(h=0\) to get the value of the limit...
Click to expand...
IMG_20190318_233433.jpg
 
Jomo said:
Please note that h/h is NOT 1. h/h is actually a piecewise function. In fact h/h = 1 if h is not 0
And h/h is indetermined if h=0
Since we have lim as h goes to 0, we never have h=0. So h/h = 1 or you can say it cancels out.
Click to expand...
Thank you very much for your help
 
MarkFL said:
It's not necessary to expand the binomial in the denominator...I wind up with (I factored the -1 out front too):

[MATH]f'(x)=-3\lim\frac{5x^4h+10x^3h^2+10x^2h^3+5xh^4+h^5}{h(x(x+h))^5}[/MATH]
Next, factor the numerator:

[MATH]f'(x)=-3\lim\frac{h\left(5x^4+10x^3h+10x^2h^2+5xh^3+h^4\right)}{h(x(x+h))^5}[/MATH]
Divide out the common factor:

[MATH]f'(x)=-3\lim\frac{5x^4+10x^3h+10x^2h^2+5xh^3+h^4}{(x(x+h))^5}[/MATH]
Now, you can substitute \(h=0\) to get the value of the limit...
Click to expand...
Thank you very much for your help,1 down 4 to go...but this was extremely helpful and fun.
 
You got the corrrect answer. If you need help with any of the other 4, then please post the problem in a new thread.
 
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