Hello All,

The following is a paragraph from my book...

"Find the slope of the tangent to the curve $\displaystyle y=\frac{1}{2x}+3$ at the point where $\displaystyle x=-1$. Find the angle at which this tangent makes with the curve $\displaystyle y=2x^2+2$. The slope of the tangent is the slope of the curve at the point where they touch one another; that is, it is the $\displaystyle \frac{dy}{dx}$ of the curve for that point. Here $\displaystyle \frac{dy}{dx}= -\frac{1}{2x^2}$ and for $\displaystyle x=-1, \frac{dy}{dx}=\frac{-1}{2}$, which is the slope of the tangent and of the curve at that point. The tangent, being a straight line, has for equation $\displaystyle y=ax+b$, and its slope is $\displaystyle \frac{dy}{dx}=a$, hence $\displaystyle a=\frac{-1}{2}$. Also, if $\displaystyle x=-1, y=\frac{1}{(2)(-1)}+3=2\frac{1}{2}$, and as the tangent passes by this point, the coordinates of that point must satisfy the equation of the tangent, namely: $\displaystyle y=-\frac{1}{2}x+b$ so that $\displaystyle 2\frac{1}{2}=\frac{-1}{2}(-1)+b$ and $\displaystyle b=2$; the equation of the tangent is therefore $\displaystyle y=\frac{-1}{2}x+2$ "

So here are my questions regarding this paragraph:

1.) When I take the tangent of the first curve, I get $\displaystyle \frac{dy}{dx}=\frac{1}{2}$ ...how does he come up with $\displaystyle \frac{dy}{dx}=\frac{1}{2x^2}$ ?

(depending on the answer, I might have more questions to follow) Thanks.