Question

If y = f(x) is a differentiable function of x on an interval I and y is one-one onto and `(dy)/(dx) ≠ 0` on I, then prove that `(dx)/(dy) = 1/(((dy)/(dx)))`, where `(dy)/(dx) ≠ 0`. Hence, prove that `d/dx (cot^-1 x) = -1/(1 + x^2)`.

Answer 1

Given that y = f(x) and x = f⁻¹(y) are differentiable functions.

Let δy be the increment in y corresponding to an increment δx in x.

∴ as δx → 0, δy → 0.

Now, y is a differentiable function of x.

∴ `lim_(deltax ->0) (deltay)/(deltax) = (dy)/(dx)`

Now, `(deltay)/(deltax) xx (deltax)/(deltay) = 1`

∴ `(deltax)/(deltay) = 1/(((deltay)/(deltax)))`

Taking limits on both sides as δx → 0, we get

`lim_(deltax ->0) (deltax)/(deltay) = lim_(deltax -> 0) [1/(((deltay)/(deltax)))] = 1/(lim_(deltax->0)(deltay)/(deltax))`

∴ `lim_(deltay->0) (deltax)/(deltay) = 1/(lim_(deltax->0)(deltay)/(deltax))        ...["as"  deltax -> 0, deltay -> 0]`

Since limit in RHS exists, limit in LHS also exists, and we have

`lim_(deltay->0) (deltax)/(deltay) = (dx)/(dy)`

∴ `(dx)/(dy) = 1/((dy"/"dx))`, where `(dy)/(dx) ≠ 0`.

To prove that `d/dx (cot^-1 x) = -1/(1 + x^2)`:

Let y = cot⁻¹ x

Then x = cot y, where x ∈ R and 0 < y < π.

∴ `(dx)/(dy) = d/dy (cot y) = -"cosec"^2y= -(1 + cot^2y) = -(1 + x^2)` 

∴ `(dy)/(dx) = 1/((dx"/"dy))`, if `(dx)/(dy) ≠ 0`

∴ `(dy)/(dx) = 1/(-(1 +x^2))`

∴ `d/(dx) (cot^-1 x) = -1/(1 + x^2)`