Question

Suppose y = f(x) is differentiable function of x and y is one-one onto, `dy/dx ≠ 0`. Also, if x = f^–1(y) is differentiable, then prove that `dx/dy = 1/((dy/dx))`, where `dy/dx ≠ 0`

Hence, find `d/dx(tan^-1x)`.

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_(δx -> 0) (δy)/(δx) = dy/dx`

Now, `(δy)/(δx) xx (δx)/(δy)` = 1

∴ `(δx)/(δy) = 1/(((δy)/(δx)))`

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

`lim_(δx -> 0) (δx)/(δy) = lim_(δx -> 0) [1/(((δy)/(δx)))] = 1/(lim_(δx -> 0) (δy)/(δx)`

∴ `lim_(δy -> 0) (δx)/(δy) = 1/(lim_(δx -> 0) (δy)/(δx)`   ...[As δx `->` 0, δy `-> 0`]

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

`lim_(δx -> 0) (δx)/(δy) = dx/dy`

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

To find `bb(d/dx(tan^-1x))`:

Let y = tan^–1x

Then x = tan y, where x ∈ R and `-π/2 < y < π/2`

Differentiating w.r.t. y, we get

`dx/dy = d/dy (tany)` = sec²y = 1 + tan²y = 1 + x²

∴ `dy/dx = 1/((dx/dy)), if dx/dy ≠ 0`

∴ `dy/dx = 1/(1 + x^2)`

∴ `d/dx (tan^-1x) = 1/(1 + x^2)`