Question

Find the equation of the tangent to the curve
tan^-1(3x) + tan^-1(3y) = Pi/2 at the point where x=1/3 Find the equation of the tangent to the curve
tan^-1(3x) + tan^-1(3y) = Pi/2 at the point where x=1/3 @Mathematics

Answer 1

derivative is
\[\frac{3}{9x^2+1}+\frac{3}{9y^2+1}y'=0\] solve for y' and plug in the numbers

Answer 2

first what is y' given
\[y=\tan^{-1}(g(x)) => \tan(y)=g(x)=\frac{g(x)}{1} (=\frac{opp}{adj})\]

\[\sec^2(y)y'=g'(x) => y'=\frac{g'(x)}{\sec^2(y)}=\frac{g'(x)}{1+g^2}\]