Question

find the equation of the tangent line to the graph of the function at the given point
y=arctan 2/x, (2, pi/4)

Answer 1

well just differentiate it then find the gradient m since the x value is given

then y -y_1 = m(x-x_1) i.e eqn. of tangent line

Answer 2

\[f(x)=\arctan(\frac{2}{x})\]
\[f'(x)=\frac{-2}{x^2+4}\]
\[f'(2)=-\frac{1}{4}\] then point - slope

Answer 3

or... :)\[tan(y)=\frac{2}{x}\]
\[sec^2(y)y'=-\frac{2}{x^2}\]
\[y'=-\frac{2cos^2(y)}{x^2}\]

Answer 4

\[y'=-\frac{2~cos^2(45^o)}{4}\]
\[y'=-\frac{cos^2(45^o)}{2}\]
\[y'=-\frac{1}{(\sqrt{2})^22}\]
\[y'=-\frac{1}{4}\]

lol