# Thread: Derivates and their graphs

1. ## Derivates and their graphs

Ok so I was wondering if anyone could help me out and check my work? i need to find the first and second derivatives of f(x) and I believe I am right but a kid that is in my class with me are having trouble checking our answer. Our teacher taught us how to do it by checking the graphs but we aren't sure that it's working out properly. Any help would be greatly appreciated.

$\displaystyle f(x)= 2x /{x}^{2} + 1$
$\displaystyle f(x)= ({x}^{2} + 1)(2)-(2x)(2x)/({{x}^{2} + 1})^{2 }$ found derivative
$\displaystyle f(x)= ({2x}^{2} + 2 - {4x}^{2 }) /({{x}^{2} + 1})^{2 }$then simplified it
$\displaystyle f(x)= ({-2x}^{2}+2)/ ({{x}^{2} + 1})^{2 }$
that is what I got for my first derivative
and this is my second:

$\displaystyle f(x)= ({{x}^{2} + 1})^{2 }(-4x)-({-2x}^{2} + 2)(2({x}^{2} + 1})(2x)/ {({{x}^{2} + 1})^{2 }}^{ 2}$ second derivative
$\displaystyle f(x)=({-4x}^{ 5}-{8x}^{3 }-4x)+({8x}^{5 }+{8x}^{ 3}-{8x}^{3 }- 8x)/( {{x}^{ 2}+1 })^{4 }$ simplified
$\displaystyle f(x)=({4x}^{5 }- {8x}^{ 3} - 4x)/ ( {{x}^{ 2}+1 })^{4 }$ simplifying some more
$\displaystyle f(x)= (4x({x}^{ 4}-{2x}^{ 2}-1))/ ( {{x}^{ 2}+1 })^{4 }$
and this is what I got for my second derivative

I was wondering if anyone could help me and tell me where I went wrong if I went wrong, or if this is completely correct. Also was wondering if anyone knew how the graphs of the first and second derivative compare to the original equation and if you could explain to me or point me in the right direction for info on that.

2. Originally Posted by eddieg0304
Ok so I was wondering if anyone could help me out and check my work? i need to find the first and second derivatives of f(x) and I believe I am right but a kid that is in my class with me are having trouble checking our answer. Our teacher taught us how to do it by checking the graphs but we aren't sure that it's working out properly. Any help would be greatly appreciated.

$\displaystyle f(x)= 2x /{x}^{2} + 1$
$\displaystyle f(x)= ({x}^{2} + 1)(2)-(2x)(2x)/({{x}^{2} + 1})^{2 }$ found derivative
$\displaystyle f(x)= ({2x}^{2} + 2 - {4x}^{2 }) /({{x}^{2} + 1})^{2 }$then simplified it
$\displaystyle f(x)= ({-2x}^{2}+2)/ ({{x}^{2} + 1})^{2 }$
that is what I got for my first derivative
and this is my second:

$\displaystyle f(x)= ({{x}^{2} + 1})^{2 }(-4x)-({-2x}^{2} + 2)(2({x}^{2} + 1})(2x)/ {({{x}^{2} + 1})^{2 }}^{ 2}$ second derivative
$\displaystyle f(x)=({-4x}^{ 5}-{8x}^{3 }-4x)+({8x}^{5 }+{8x}^{ 3}-{8x}^{3 }- 8x)/( {{x}^{ 2}+1 })^{4 }$ simplified
$\displaystyle f(x)=({4x}^{5 }- {8x}^{ 3} - 4x)/ ( {{x}^{ 2}+1 })^{4 }$ simplifying some more
$\displaystyle f(x)= (4x({x}^{ 4}-{2x}^{ 2}-1))/ ( {{x}^{ 2}+1 })^{4 }$
and this is what I got for my second derivative

I was wondering if anyone could help me and tell me where I went wrong if I went wrong, or if this is completely correct. Also was wondering if anyone knew how the graphs of the first and second derivative compare to the original equation and if you could explain to me or point me in the right direction for info on that.
First note that the LaTex command \frac{}{} will make your posts easier to read.

Your first derivative is correct as is the first line of your 2nd derivative.

$\displaystyle f''(x)= \frac{(x^{2} + 1)^{2}(-4x)-(-2x^{2} + 2)(2(x^{2} + 1)(2x))}{(x^2 + 1)^4 }$

From here factor the numerator to get

$\displaystyle f''(x)= \frac{4x(x^2+1)[(x^{2} + 1)(-1)-(-2x^{2} + 2))]}{(x^2 + 1)^4 }$

Now collect like terms and simplify

$\displaystyle f''(x)= \frac{4x[x^3-3]}{(x^2 + 1)^3 }$

As for the 2nd part I am not quite sure what you are asking. The first derivative tells you the slope of the original function and the 2nd derivative tells you about the concavity of the function.

3. Originally Posted by TheEmptySet
First note that the LaTex command \frac{}{} will make your posts easier to read.

Your first derivative is correct as is the first line of your 2nd derivative.

$\displaystyle f''(x)= \frac{(x^{2} + 1)^{2}(-4x)-(-2x^{2} + 2)(2(x^{2} + 1)(2x))}{(x^2 + 1)^4 }$

From here factor the numerator to get

$\displaystyle f''(x)= \frac{4x(x^2+1)[(x^{2} + 1)(-1)-(-2x^{2} + 2))]}{(x^2 + 1)^4 }$

Now collect like terms and simplify

$\displaystyle f''(x)= \frac{4x[x^3-3]}{(x^2 + 1)^3 }$

As for the 2nd part I am not quite sure what you are asking. The first derivative tells you the slope of the original function and the 2nd derivative tells you about the concavity of the function.
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ok very awesome... and thank you very much... and what i ment was that my teacher would explain a way to tell if the derivatives are correct by analyzing the graphs from the original equation and the first and second derivative. I believe she said that the derivatives are opposites, so when one is decreasing the other is increasing and so on. I don't know if i misunderstood how she explained it, or if that is how you can tell if the derivatives are correct from the graph

4. Originally Posted by TheEmptySet
From here factor the numerator to get

$\displaystyle f''(x)= \frac{4x(x^2+1)[(x^{2} + 1)(-1)-(-2x^{2} + 2))]}{(x^2 + 1)^4 }$
Hello TheEmptySet. I'm Eddie's classmate. Thank you for your help thus far. We both are admittedly struggling with the algebra aspects of the course / question at hand. Could you break down this factor step into what parts you pull apart / combine from the initial quotient rule?

Also, for the second portion of the question, we're having a hard time identifying critical points and points of inflection. I'm not ignoring my textbook, but I'm having a hard time seeing how those concepts apply to this function.

For example, by looking at the graphs of the functions:
$\displaystyle f(x)=2x^3, f'(x)=6x^2, f''(x)=12x$
I can easily see where critical points are located as $\displaystyle f'(x)$ could also be thought of as $\displaystyle \frac{dy}{dx}$. But this is a very simple case.

I believe breaking the complexity down into basic steps will help knock some sense .
Hopefully this helps define the questions better.