Question

What is the implicit derivative of `1=cosx/y`?

Answer 1

`dy/dx=-(ysinx)/cosx`

Explanation:

If we differentiate implicitly and use the quotient rule, we see that:

`d/dx[1=cosx/y]`

`0=(stackrel"-sinx"overbrace(d/dx[cosx])*y-cosx*stackrel"dy/dx"overbrace(d/dx[y]))/y^2`

`(-ysinx-cosx[dy/dx])/y^2=0`

`-ysinx-cosx[dy/dx]=0`

`dy/dx=-(ysinx)/cosx`