MHB Help with finding Rate of change of f

[riri]

Hi! This is my first post and I'm hoping to receive so help :D
If f(x)=\frac{1}{{x}^{2}+1}. The rate of change of f at a is defined as \lim_{{h}\to{0}}\frac{f(a+h)-f(a)}{h}.
In case, these commands don't work properly... the question is: if f(x)=1/(x^2+1) and rate of change of f at a is defined as lim h-->0 (f(a+h)-f(a))/h, what would rate of change of f at 2 be?

I tried to solve it, but the result was either 0 or undefined, which is wrong, so can anyone help on how to solve this?
Thanks!

 

[MarkFL]

Hello and welcome to MHB! :D

Okay, using the definitions:

$$f'(a)\equiv\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$

and:

$$f(x)=\frac{1}{x^2+1}$$

Find:

$$f'(2)$$

Okay, well first using the definition of the derivative and the given function, we may write:

$$f'(a)=\lim_{h\to0}\frac{\dfrac{1}{(a+h)^2+1}-\dfrac{1}{a^2+1}}{h}$$

The next thing we want to do is combine the terms in the numerator:

$$\dfrac{1}{(a+h)^2+1}-\dfrac{1}{a^2+1}$$

What do you get when you perform the subtraction?

Note: To use $\LaTeX$, click the $\Sigma$ button on the bottom row of our editor toolbar, and this will generate MATH tags, inside of which you can put your code. :)

 

[riri]

Hi and thank you so much :D

I did what you outlined, and I'm a bit confused with the subtraction part... is it \frac{1}{2ah+h^2} ?
In terms of the actual numbers, I got up to:
(\lim_{{h}\to{0}}\frac{1}{4+4h+h^2+1}-\lim_{{h}\to{0}}\frac{1}{5})/h

And then i just did what you said, and I got \frac{1}{h^2+4h} and then factored it out, \frac{1}{h(h+4)} and then substituted the 0 into it, but that's wrong so I'm just lost somewhere in between

Thank you so much!

 

[MarkFL]

I would derive the result first before plugging in any numbers, so let's go back to:

$$\frac{1}{(a+h)^2+1}-\frac{1}{a^2+1}$$

Performing the subtraction, we obtain:

$$\frac{\left(a^2+1\right)-\left((a+h)^2+1\right)}{\left((a+h)^2+1\right)\left(a^2+1\right)}$$

If we simplify the numerator, we have:

$$\frac{a^2-(a+h)^2}{\left((a+h)^2+1\right)\left(a^2+1\right)}$$

Now, what do you get if you apply the difference of squares formula to the numerator?

 

[riri]

Okay, so to apply difference of squares would I expand the a^2-(a+h)^2 first?
So this would lead to a^2-(a^2+2ah+h^2)?
Then do I apply difference of squares to (a^2+2ah+h^2? which...probably isn't right either so
It wouldn't be right to eliminate the a^2 since a^2-a^2-2ah-h^2?

 

[MarkFL]

Okay, so to apply difference of squares would I expand the a^2-(a+h)^2 first?
So this would lead to a^2-(a^2+2ah+h^2)?
Then do I apply difference of squares to (a^2+2ah+h^2? which...probably isn't right either so
It wouldn't be right to eliminate the a^2 since a^2-a^2-2ah-h^2?

I left the numerator as a difference of squares, so we could then write:

$$\frac{a^2-(a+h)^2}{\left((a+h)^2+1\right)\left(a^2+1\right)}=\frac{(a+(a+h))(a-(a+h))}{\left((a+h)^2+1\right)\left(a^2+1\right)}=\frac{(2a+h)(-h)}{\left((a+h)^2+1\right)\left(a^2+1\right)}=-\frac{h(2a+h)}{\left((a+h)^2+1\right)\left(a^2+1\right)}$$

And so the derivative becomes:

$$f'(a)=\lim_{h\to0}-\frac{h(2a+h)}{h\left((a+h)^2+1\right)\left(a^2+1\right)}$$

Can you proceed?

 

[riri]

Sorry for the ultra late reply!
Thank you so much, I understand after reviewing more for a while :)
So would the answer just be \frac{h(4+2h)}{5(h+4h+5)}?
Or would I leave it at \frac{4h+2h^2}{5h^2+20h+25}?Thank you so much!

 

[MarkFL]

Sorry for the ultra late reply!
Thank you so much, I understand after reviewing more for a while :)
So would the answer just be \frac{h(4+2h)}{5(h+4h+5)}?
Or would I leave it at \frac{4h+2h^2}{5h^2+20h+25}?Thank you so much!

After we divide out common factors, we have:

$$f'(a)=-\lim_{h\to0}\frac{2a+h}{\left((a+h)^2+1\right)\left(a^2+1\right)}$$

Now we no longer have the indeterminate form 0/0, and so we can find the limit by direct substitution for $h$:

$$f'(a)=-\lim_{h\to0}\frac{2a+h}{\left((a+h)^2+1\right)\left(a^2+1\right)}=-\frac{2a+0}{\left((a+0)^2+1\right)\left(a^2+1\right)}=-\frac{2a}{\left(a^2+1\right)^2}$$

Now we are ready to let $a=2$, so that we find:

$$f'(2)=-\frac{2(2)}{\left(2^2+1\right)^2}=-\frac{4}{25}$$