Question

How do you implicitly differentiate `-3=xsecy`?

Answer 1

`(dy)/(dx)=-xcoty`

Explanation:

When we implicitly differentiate a function `f(y)`, we first differentiate `f(y)` with respect to `y` and then multiply by `(dy)/(dx)`. This comes from chain rule as `(df)/(dx)=(df)/(dy)xx(dy)/(dx)`.

Now using product rule for differentiating `3=xsecy`, we get

`0=x xxsecy+1xxsecytany(dy)/(dx)` or

`secytany(dy)/(dx)=-xsecy` or

`(dy)/(dx)=(-xsecy)/(secytany)`

= `(-xcancel(secy))/(cancel(secy)tany)`

= `-xcoty`