Using Derivatives and Tangent Line to Find Area Let $(a, b)$ be an arbitrary point on the graph of $y=\frac1x$ ($x>0$). Prove that the area of the triangle formed by the tangent line through $(a,b)$ and the coordinate axes is $2$ square units.
I know that I need to use derivatives and I've already come up with $f'(a)=\frac{1}{2a^2}$, but I am not sure how to use that or where to go from there.
Thanks!

First, we compute the derivative of the function $ f(x) = \frac{1}{x} $, which is actually $ -\frac{1}{x^2} $. (If you want to know more about this, try looking up "Quotient Rule" in Google!)
So the tangent line at $ (a,b) = (a, \frac{1}{a}) $ should satisfy the following:


*

*Its slope must be $ - \frac{1}{a^2} $ at all values of x.

*It must pass through $ (a, \frac{1}{a}) $.


With those conditions in mind, we can set up the equation of the line: $ y=-\frac{1}{a^2}(x-a)+\frac{1}{a} $. Now compute the x-intercept and y-intercept. Those are $ 2a \text{ and } \frac{2}{a} $, respectively. Therefore the area of the triangle equals to $ 2a \times \frac{2}{a} \times \frac{1}{2} $, which is 2 units.
Hope this helped!
