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!
$\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.