Question

How do you differentiate `p(y) = e^(y^2)+sin^2(y)cos(y)-ye^(y)` using the product rule?

Answer 1

`(dp)/(dy)=2ye^(y^2)+2sinycos^2y-sin^3y-e^y-ye^y`

Explanation:

`p(y)=e^(y^2)+sin^2ycosy-ye^y`

Hence, `(dp)/(dy)=e^(y^2)xx2y+(2sinycosyxxcosy+sin^2yxx(-siny))-(1xxe^y+ye^y)`

= `2ye^(y^2)+2sinycos^2y-sin^3y-e^y-ye^y`