Solving an Algebraic expression to get x

[Naz]

Hello,

Been stuck all day with this problem an wondered if anyone can help.

It comes from a differentiation chain rule question. I've done the differentiation bit, but can't figure out how to do the algebra at the end

The question is find dy/dx when
y= 1/(x²-9) and then find the x co-ordinate of the stationary point.

I get the derivative to be
= -2x+9
(x²-9x)<sup>2

However that's as far as I can go as I do not know how to solve for x when the derivative =0
Can someone show me how to do this?

Thanks</sup>

 

[srmichael]

Hello,

Been stuck all day with this problem an wondered if anyone can help.

It comes from a differentiation chain rule question. I've done the differentiation bit, but can't figure out how to do the algebra at the end

The question is find dy/dx when
y= 1/(x²-9) and then find the x co-ordinate of the stationary point.

I get the derivative to be
= -2x+9
(x²-9x)<sup>2

However that's as far as I can go as I do not know how to solve for x when the derivative =0
Can someone show me how to do this?

Thanks</sup>

First, there is a typo in the original equation you wrote. It should have said \(\displaystyle y=\dfrac{1}{x^2-9x}\)

That being said, the only way a fraction can equal 0 is when the numerator equals 0. Can you take it from there?

 

[Naz]

First, there is a typo in the original equation you wrote. It should have said \(\displaystyle y=\dfrac{1}{x^2-9x}\)

That being said, the only way a fraction can equal 0 is when the numerator equals 0. Can you take it from there?

Thank you very much.

Seems so obvious now.

 

[HallsofIvy]

Hello,

Been stuck all day with this problem an wondered if anyone can help.

It comes from a differentiation chain rule question. I've done the differentiation bit

No, you've don't the "differentiation bit" incorrectly, but can't figure out how to do the algebra at the end

The question is find dy/dx when
y= 1/(x²-9) and then find the x co-ordinate of the stationary point.

I get the derivative to be
= -2x+9
(x²-9x)²[/quote]
The "quotient rule" says that the derivative of \(\displaystyle \frac{u(x)}{v(x)}\) is \(\displaystyle \frac{u'v- uv'}{v^2}\).
Here u(x)= 1 so that u'(x)= 0 and \(\displaystyle v(x)= x^2- 9\) so that v'= 2x (NOT "2x- 9"- that may be your error)
So the derivative is \(\displaystyle \frac{0(x^2- 9)- (1)(2x)}{(x^2- 9)}= \frac{-2x}{(x^2- 9)}\)

You mention the "differentiation chain rule". Instead of the quotient rule, you can use the chain rule by writing the function as \(\displaystyle y= (x^2- 9)^{1}\) so that \(\displaystyle y'= -(x^2- 9)^{-2}(2x)= \frac{-2x}{(x^2- 9)^2}\).

However that's as far as I can go as I do not know how to solve for x when the derivative =0
Can someone show me how to do this?

Thanks
[/FONT]

A fraction is 0 if and only if the numerator is 0.