Question

How do you differentiate `y=sinx+x^2tan^-1x`?

Answer 1

`y'(x)=cos(x)+2xarctan(x)+x^2/(1+x^2)`

Explanation:

Note that
`(arctan(x))'=1/(1+x^2)`
and by the sum and product rule we get

`y'(x)=cos(x)+2xarctan(x)+x^2/(1+x^2)`

Answer 2

`dy/dx=cosx+x^2/(1+x^2)+2xtan^-1x`

Explanation:

`"differentiate "x^2tan^-1x" using the "color(blue)"product rule"`

`"given "y=f(x)g(x)" then"`

`dy/dx=f(x)g'(x)+g(x)f'(x)larrcolor(blue)"product rule"`

`f(x)=x^2rArrf'(x)=2x`

`g(x)=tan^-1xrArrg'(x)=1/(1+x^2)`

`d/dx(x^2tan^-1x)`

`=x^2/(1+x^2)+2xtan^-1x`

`dy/dx=cosx+x^2/(1+x^2)+2xtan^-1x`