Derieving Derivative #4

Taking $\frac{dy}{dx} = -\frac{c^2}{x^2}$ , the tangent line at the arbitrary point $(x_0, \frac{c^2}{x_0})$ is given by:

$y - \frac{c^2}{x_0} = (-\frac{c^2}{x_0^{2}})(x-x_0) \Rightarrow y = -\frac{c^2}{x_0^{2}} x + \frac{2c^2}{x_0}$

with $x$ and $y-$ intercepts at $(2x_0,0), (0,\frac{2c^2}{x_0})$ respectively. The resultant right triangle formed between the tangent line and the coordinate axes yields an area of:

$A = \frac{1}{2} \cdot (2x_0)(\frac{2c^2}{x_0}) = \boxed{2c^2}.$