Question

What is the implicit derivative of `1=tanyx-y^2`?

Answer 1

`dy/dx=(ysec^2(yx))/(2y-xsec^2(yx))`

Explanation:

`"differentiate "color(blue)"implicitly with respect to x"`

`"differentiate "tan(yx)" using "color(blue)"chain/product rules"`

`0=sec^2(yx)xxd/dx(yx)-2ydy/dx`

`rArr2ydy/dx=sec^2(yx)(y+xdy/dx)`

`color(white)(rArr2ydy/dx)=ysec^2(yx)+xsec^2(yx)dy/dx`

`rArr2ydy/dx-xsec^2(yx)dy/dx=ysec^2(yx)`

`rArrdy/dx(2y-xsec^2(yx))=ysec^2(yx)`

`rArrdy/dx=(ysec^2(yx))/(2y-xsec^2(yx))`