Question

How do you find `(d^2y)/(dx^2)` given `x=tany`?

Answer 1

`AA y in RR, (d^2y)/dx^2=(-2x)/(1+x^2)^2`.

Explanation:

`x=tan y`.

Case : 1 `"Suppose that,"y in (-pi/2,pi/2) rArr y=arc tan x, x in RR`

`:. dy/dx=1/(1+x^2)`.

`:. (d^2y)/dx^2=d/dx(dy/dx)=d/dx(1/(1+x^2))`

Since, `d/dt(1/t)=-1/t^2`, we have, by the Chain Rule,

`(d^2y)/dx^2=-1/((1+x^2)^2)*d/dx(1+x^2)=(-2x)/((1+x^2)^2)`

Case : 2 `"Next, let "y in (pi/2,3pi/2) rArr pi/2<,y<,3pi/2`

`rArr pi/2-pi<,y-pi<,3pi/2-pi`

`rArr -pi/2<,y-pi<,pi/2, &, tan(y-pi)=-tan(pi-y)=-(-tany)=tany=x`.

Thus, in this Case, `x=tan(y-pi), where, (y-pi) in (-pi/2,pi/2)`

`:." by defn. of arc tan, "y-pi=arc tanx, or, y=pi+arc tan x`

`:. (d^2y)/dx^2=(-2x)/(1+x^2)^2`, as in Case : 1

Thus, `AA y in RR, (d^2y)/dx^2=(-2x)/(1+x^2)^2`.

Answer 2

To differentiate without using the inverse tangent, see below.

Explanation:

`tan y = x`

Differentiate both sides with respect to `x`.

`d/dx(tanx) = d/dx(x)`

`sec^2y dy/dx = 1`

`dy/dx = cos^2y`

Differentiate again w.r.t. `x`

`(d^2y)/dx^2 = 2cos y (-siny dy/dx)`

Now rfeplace `dy/dx`

`(d^2y)/dx^2 = -2 siny cos^3 y`

To see that this is the same as the other answer

`tany = x` `rArr` `cos y = 1/(1+x^2)`

(To see this draw and label a right triangle with angle `y`, opposite side `x` and adjacent side `1`. So the hypotenuse is `1+x^2`.)

(Or use `tany=x` `rArr` `tan^2y = x^2`, so `1+tan^2y = 1+x^2` and `sec^y = 1+x^2` so that `cos y = 1/(1+x^2)`

Now,

`(d^2y)/dx^2 = -2 siny cos^3 y`

`= -2 siny/cosy cos^4y`

`= -2tany (cos^2y)^2`

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

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