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Help on C3 Differentiation Pls

The question is: f(x)=x/x2+2 f(x) = x/x^2+2 Find the set of values of x for which f'(x)<0
(edited 6 years ago)

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Original post by AxSirlotl
The question is: f(x)=x/x2+2 f(x) = x/x^2+2 Find the set of values of x for which f'(x)<0


The first step is obviously to get an expression for f'(x), have you managed to do that?
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AxSirlotl
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Original post by Plagioclase
The first step is obviously to get an expression for f'(x), have you managed to do that?


yeah boii
Original post by AxSirlotl
The question is: f(x)=x/x2+2 f(x) = x/x^2+2 Find the set of values of x for which f'(x)<0


You have to use the quotient rule and then look at the numerator.

Could you show your workings?
(edited 6 years ago)
Original post by AxSirlotl
yeah boii


Well then, you've got your inequality, you just need to solve that.
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AxSirlotl
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Original post by BrasenoseAdm
You have to use the quotient rule and then look at the numerator.

Could you show your workings?


I got f′(x)=−2x2/x2+2 f'(x) = -2x^2/x^2+2
Sorry, I can't do proper fractions because I don't understand the guide heh
Original post by AxSirlotl
I got f′(x)=−2x2/x2+2 f'(x) = -2x^2/x^2+2
Sorry, I can't do proper fractions because I don't understand the guide heh


Your derivative is wrong. To write a fraction it's \frac{numerator}{denominator}, but the notation you're using is acceptable anyway.
Original post by NotNotBatman
Your derivative is wrong. To write a fraction it's \frac{numerator}{denominator}, but the notation you're using is acceptable anyway.


Does TSR detect LaTex automatically? We tried replying with the coding:

\frac{x}{x^2+2}

&

\frac{x}{x^2+2}
(edited 6 years ago)
Original post by BrasenoseAdm
Does TSR detect LaTex automatically? We tried replying with the coding:

\frac{x}{x^2+2}

&

\frac{x}{x^2+2}


maths things go here , but without the space after the /
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AxSirlotl
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Original post by NotNotBatman
Your derivative is wrong. To write a fraction it's \frac{numerator}{denominator}, but the notation you're using is acceptable anyway.


I keep getting the same answer ;-;

I tried doing fractions but it kept putting my numerator in front of my fraction and leaving the top of the fraction blank, so I can't really show my working very easily.
Original post by AxSirlotl
I keep getting the same answer ;-;

I tried doing fractions but it kept putting my numerator in front of my fraction and leaving the top of the fraction blank, so I can't really show my working very easily.


Thanks for the LaTex tip!

Problem: differentiate xx2+2\frac{x}{x^2+2}

Technique: quotient rule vu′−uv′v2\frac{vu' - uv'}{v^2}

u = x and v = X^2+2, u' = derivative of u and v'= derivative of v
Use the chain rule for differentiation:

h(x) = f(x)/g(x)

h'(x) = f'(x)g(x)-f(x)g'(x) / [g(x)]^2

The last step is trivial; let the inequality be less than 0 and solve it accordingly.
Is that xx2+2 \frac{x}{x^2} + 2 or xx2+2 \frac{x}{x^2+2}
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AxSirlotl
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Original post by zeldor711
Is that xx2+2 \frac{x}{x^2} + 2 or xx2+2 \frac{x}{x^2+2}


The latter
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AxSirlotl
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Original post by BrasenoseAdm
Thanks for the LaTex tip!

Problem: differentiate xx2+2\frac{x}{x^2+2}

Technique: quotient rule vu′−uv′v2\frac{vu' - uv'}{v^2}

u = x and v = X^2+2, u' = derivative of u and v'= derivative of v


Original post by thekidwhogames
Use the chain rule for differentiation:

h(x) = f(x)/g(x)

h'(x) = f'(x)g(x)-f(x)g'(x) / [g(x)]^2

The last step is trivial; let the inequality be less than 0 and solve it accordingly.


I got my derivative as -1/(x^2+2), is that right ;-;
Original post by AxSirlotl
I got my derivative as -1/(x^2+2), is that right ;-;


Not quite.

If you post your working we would just be able to spot where you're going wrong.
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AxSirlotl
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Original post by RDKGames
Not quite.

If you post your working we would just be able to spot where you're going wrong.


Using the quotient rule:

u = x v = x^2 + 2

u' = 1 v' = 2x

Then I got
numerator: (x^2 + 2) - 2x^2

denominator: (x^2 + 2)^2

So on the top I had v x u' - u x v' all over v^2
Original post by AxSirlotl
Using the quotient rule:

u = x v = x^2 + 2

u' = 1 v' = 2x

Then I got
numerator: (x^2 + 2) - 2x^2

denominator: (x^2 + 2)^2

So on the top I had v x u' - u x v' all over v^2


OK, so how did you go from 2−x2(x2+2)2\frac{2-x^2}{(x^2+2)^2} to −1x2+2-\frac{1}{x^2+2}?
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AxSirlotl
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Original post by RDKGames
OK, so how did you go from 2−x2(x2+2)2\frac{2-x^2}{(x^2+2)^2} to −1x2+2-\frac{1}{x^2+2}?


I've just realised that's wrong, I mean I've had another go at it and I'm back to (-x^2 + 2)/(x^2+2)^2
I'm not sure where to go from there, if I need make that into an inequality then I'm not sure what to do after that.
Original post by AxSirlotl
I've just realised that's wrong, I mean I've had another go at it and I'm back to (-x^2 + 2)/(x^2+2)^2
I'm not sure where to go from there, if I need make that into an inequality then I'm not sure what to do after that.


(x2+2)2>0(x^2+2)^2 > 0 so 2−x2(x2+2)2<0⇒2−x2<0\frac{2-x^2}{(x^2+2)^2} <0 \Rightarrow 2-x^2 <0

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