## anonymous one year ago Need help implicit differentiation Tan(x+y)=x

1. anonymous

$\tan \left( x+y \right)=x$ diff. w.r.t x $\sec ^2\left( x+y \right)\left\{ 1+\frac{ dy }{ dx } \right\}=1$ $1+\frac{ dy }{ dx }=\cos ^2\left( x+y \right)$ $\frac{ dy }{ dx }=\cos ^2\left( x+y \right)-1=-\sin ^2\left( x+y \right)$

2. anonymous

Thanks ! I need to revisit my trig iD's