Answer:
[tex]\dfrac{\text{d}}{\text{d}x}\left(\arctan\left(\dfrac{1}{x}\right)\right)=-\dfrac{1}{1+x^2}[/tex]
Step-by-step explanation:
Given:
[tex]y = \arctan\left(\dfrac{1}{x}\right)[/tex]
Take the tangent of both sides:
[tex]\tan y = \dfrac{1}{x}[/tex]
Differentiate both sides with respect to x:
[tex]\sec^2y \;\dfrac{\text{d}y}{\text{d}x}=-\dfrac{1}{x^2}[/tex]
Note that we used the chain rule to differentiate tan y with respect to x. In practice, this means differentiate with respect to y, and place dy/dx at the end.
Rearrange to isolate dy/dx:
[tex]\dfrac{\text{d}y}{\text{d}x}=-\dfrac{1}{x^2\cdot \sec^2y}[/tex]
Now, use one of the trigonometric identities, sec²y = tan²y + 1, to rewrite sec²y in terms of tan y:
[tex]\dfrac{\text{d}y}{\text{d}x}=-\dfrac{1}{x^2(\tan^2y+1)}[/tex]
[tex]\textsf{As}\; \tan y = \dfrac{1}{x}, \textsf{ then }\tan^2y=\dfrac{1}{x^2}. \;\;\textsf{Substitute this in:}[/tex]
[tex]\dfrac{\text{d}y}{\text{d}x}=-\dfrac{1}{x^2\left(\dfrac{1}{x^2}+1\right)}[/tex]
Simplify:
[tex]\dfrac{\text{d}y}{\text{d}x}=-\dfrac{1}{\dfrac{x^2}{x^2}+x^2} \\\\\\ \dfrac{\text{d}y}{\text{d}x}=-\dfrac{1}{1+x^2}[/tex]
Therefore, we have proved that the derivative of arctan(1/x) is:
[tex]\dfrac{\text{d}}{\text{d}x}\left(\arctan\left(\dfrac{1}{x}\right)\right)=-\dfrac{1}{1+x^2}[/tex]
[tex]\dotfill[/tex]
Additional Notes
Your errors lie in your definition of tan(θ), the labelling of the sides of your right triangle, and your initial statement.
The tangent ratio is the ratio of the side opposite the angle to the side adjacent the angle. Therefore, in a right triangle, the side opposite angle θ should be labelled '1', the side adjacent angle θ should be labelled 'x', and the hypotenuse should be labelled [tex]\sqrt{1+x^2}[/tex] (using the Pythagorean Theorem).
Therefore:
[tex]\tan \theta = \dfrac{1}{x} \implies \theta = \arctan \left( \dfrac{1}{x}\right)[/tex]
Additionally, the cosine ratio is the ratio of the side adjacent to the angle to the hypotenuse of the right triangle. Therefore:
[tex]\cos \theta=\dfrac{x}{\sqrt{1+x^2}}[/tex]
This means that:
[tex]\cos^2 \theta=\left(\dfrac{x}{\sqrt{1+x^2}}\right)^2=\dfrac{x^2}{1+x^2}[/tex]
As the secant of an angle is the reciprocal of the cosine of that angle, then:
[tex]\sec^2 \theta=\dfrac{1+x^2}{x^2}[/tex]
