Find the differential dy of the following functions

Question

a) y = tan^-1(x-√(1+x²))

b) cot^-1(√x) + cot^-1(1/x)

Answer 1

For these, you use the chain rule:

if u is a function of x, and you have y = f(u), then:

dy/dx = dy/du * du/dx

First problem:

y = tan^-1(x-√(1+x²))

u = x-√(1+x²) = x - (1+x² )^1/2

The second term will also use the chain rule, so we can use v = 1 + x², and dv/dx = 2x

du/dx = 1 - 1/2 (1+x²)^-1/2 * 2x = 1 - x(1+x²)^-1/2

So:

y = tan^-1(u)

dy/du = 1/(u² + 1)

and dy/dx = 1/(u² + 1) * 1 - x(1+x²)^-1/2

replace u to get:

dy/dx = 1/( x - (1+x² )^1/2)² + 1) * 1 - x(1+x²)^-1/2

expand and simplify as needed....

For the 2nd problem:

y = cot^-1(√x) + cot^-1(1/x)

use u (x)^1/2 and v = x^-1

giving: du/dx = 1/2x^-1/2 and dv/dx = -x^-2

and:

y = cot^-1 u + cot^-1 v

and dy/dx = dy/du (cot^-1 u) * du/dx + dy/dv(cot^-1 v) * dv/dx

to finish find dy/du and dy/dv, then replace u, v, du/dx, and du/dv and simplify as needed