Can someone show me how to find the value of the derivative of the function at the given point :

$\displaystyle g(x)=\frac{4x-5}{x^2-1}$ at the point $\displaystyle (0,5)$

Thank you!:)

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- May 29th 2009, 04:15 PMhemiDerivative value and point finding
Can someone show me how to find the value of the derivative of the function at the given point :

$\displaystyle g(x)=\frac{4x-5}{x^2-1}$ at the point $\displaystyle (0,5)$

Thank you!:) - May 29th 2009, 04:21 PMSpec
Use the quotient rule.

If $\displaystyle y(x)=\frac{g(x)}{h(x)}$ then $\displaystyle y'(x)=\frac{g'(x)h(x)-g(x)h'(x)}{[h(x)]^2}$ - May 29th 2009, 07:46 PMhemi
I really don't mean to be a pest, but could you go a bit further and show me how I'g plug in these variables. All this is pretty new to me (I know that's a terrible excuse) ... Sorry. :/ I'm having a lot of trouble with this stuff. (Worried)

- May 30th 2009, 02:43 AMSpec
$\displaystyle y(x)=\frac{4x-5}{x^2-1}

$

Let $\displaystyle g(x)=4x-5$ and $\displaystyle h(x)=x^2-1$

Then $\displaystyle g'(x)=4$ and $\displaystyle h'(x)=2x$

Plug this into the formula above and you get:

$\displaystyle y'(x)=\frac{g'(x)h(x)-g(x)h'(x)}{[h(x)]^2}=\frac{4(x^2-1)-2x(4x-5)}{(x^2-1)^2}=...$

At the point (0,5), the x value is 0, so just plug in 0 instead of x into the derivative.