Question

Using implicit differentiation, we can obtain the tangent line to the curve sin x+cos y=1 at point (π/2,π/2), i. e. =?

Answer 1

`y=pi/2`

Explanation:

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

`cosx-sinydy/dx=0`

`-sinydy/dx=-cosx`

`dy/dx=cosx/sinylarr"gradient of tangent"`

`"at "(pi/2,pi/2)`

`dy/dx=cos(pi/2)/(sin(pi/2))=0/1=0`

`"this indicates the line is horizontal with equation"`

`y=pi/2`