Question

implicity
4cosxsiny=1

Answer 1

\[4\cos(x)\sin(y)=1\]\[\cos(x)\sin(y)=1/4\]\[d/dx(\cos(x)\sin(y))=d/dx(1/4)\]\[d/dx(\cos(x)\sin(y))=0\]\[-\sin(x)\sin(y)+\cos(x)\cos(y)dy/dx=0\]
To get to the above equation I used product rule and to derive sin(y) I used chain rule, i.e. derivative of sin with y plugged in times the derivative of y.\[dy/dx=\sin(x)\sin(y)/\cos(x)\cos(y)=\tan(x)\tan(y)\]

Answer 2

ok, so if i have a numer in front of sin x i just move it to the other side?