Comment on Activity 6

`d/(dx) tan y`	`=d/(dx) x`
`sec^2 y (dy)/(dx)`	`=1`
`(dy)/(dx)`	`=1/(sec^2 y)`
`(dy)/(dx)`	`=1/(1+tan^2 y)`
`(dy)/(dx)`	`=1/(1+x^2)`

We first apply implicit differentiation to the formula `tan y=x`, where `y=tan^(-1) x:`

`(dy)/(dx)`	`=1/((dx)/(dy))`
	`=1/(d/(dy) tan y)`
	`=1/(sec^2 y)`
	`=1/(1+tan^2 y)`
	`=1/(1+x^2)`

Thus, `d/(dx) tan^(-1) x=1/(1+x^2)`. Now we see that the inverse function formula leads to the same result:

It's no surprise that these calculations are so similar, since the inverse function formula was derived by applying the Chain Rule to the defining formula for inverse functions.