Question

Find dy/dx by implicit differentiation.
y cos x = 2x^2 + 3y^2

Answer 1

when you differentiate every term attached to y, put dy/dx.
\[ycosx=2x^2+3y^2\]\[y(-sinx)+cosx(1)\frac{ dy }{ dx }=4x+6y \frac{ dy }{ dx }\]\[-ysinx+cosx \frac{ dy }{ dx }=4x+6y \frac{ dy }{ dx}\]
Now make dy/dx the subject:\[cosx \frac{ dy }{ dx }-6y \frac{ dy }{ dx }=4x+ysinx\]\[\frac{ dy }{ dx }(cosx-6y)=4x+ysinx\]\[\frac{ dy }{ dx }=\frac{ 4x+ysinx }{ cosx-6y }\]