Activity 13 Comments

Comment on Activity 7

The third side of the triangle in Figure 4 is `sqrt(1-x^2)`, so this is also `cos y`.

`d/(dx) sin y`
`=d/(dx) x`

`cos y (dy)/(dx)`
`=1`

`(dy)/(dx)`
`=1/(cos y)`

`(dy)/(dx)`
`=1/sqrt(1-x^2)`

We differentiate implicitly the formula `sin y=x`, where `y=sin^(-1) x:`

`(dy)/(dx)`
`=1/((dx)/(dy))`

`=1/(d/(dy) sin y)`

`=1/(cos y)`

`=1/sqrt(1-x^2)`

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

`d/(dx) sin y`	`=d/(dx) x`
`cos y (dy)/(dx)`	`=1`
`(dy)/(dx)`	`=1/(cos y)`
`(dy)/(dx)`	`=1/sqrt(1-x^2)`

`(dy)/(dx)`	`=1/((dx)/(dy))`
	`=1/(d/(dy) sin y)`
	`=1/(cos y)`
	`=1/sqrt(1-x^2)`