Differentiate tan^-1((√(1 + x^2 - 1)/x)) with respect to sin^-1(d(2x)/(1 + x^2)), if -1 < x < 1, x ≠ 0.

Question

Differentiate tan^-1\(\Big(\cfrac{\sqrt{1+\text x^2}-1}{\text x}\Big)\) with respect to sin^-1\(\Big(\cfrac{2\text x}{1+\text x^2}\Big)\), if -1 < x < 1, x ≠ 0.

Answer 1

Let u = tan^-1\(\Big(\cfrac{\sqrt{1+\text x^2}-1}{\text x}\Big)\) and v  = sin^-1\(\Big(\cfrac{2\text x}{1+\text x^2}\Big)\).

We need to differentiate u with respect to v that is find \(\cfrac{du}{dv}.\)

We have u =  tan^-1\(\Big(\cfrac{\sqrt{1+\text x^2}-1}{\text x}\Big)\)

By substituting x = tan θ, we have

Given –1 < x < 1 ⇒ x ϵ (–1, 1)

However, x = tan θ

⇒ tan θ ϵ (–1, 1)

On differentiating u with respect to x, we get

Now, we have

By substituting x = tan θ, we have

However,

Hence, v = sin^–1(sin2θ) = 2θ

⇒ v = 2tan^–1x

On differentiating v with respect to x, we get

We have

Thus,

\(\cfrac{du}{dv} = \cfrac14\)