Question

If x + y = sec-1 (x + y)then find dy/dx​

Answer 1

dy/dx = - 1

Step-by-step explanation:

Given,

x + y = sec⁻¹ (x + y)

or, x + y = sec(x + y) ..... (1)

Now, differentiating both sides with respect to x, we get

d/dx (x + y) = d/dx {sec(x + y)}

or, dx/dx + dy/dx = sec(x + y) tan(x + y) d/dx (x + y)

or, 1 + dy/dx = (x + y) tan(x + y) (1 + dy/dx) [by (1)]

or, 1 + dy/dx = (x + y) tan(x + y) + (x + y) tan(x + y) dy/dx

or, {1 - (x + y) tan(x + y)} dy/dx = (x + y) tan(x + y) - 1 [∵ 1 - (x + y) tan(x + y) ≠ 0]

or, dy/dx = - 1

Rules of derivatives:

1. d/dx (x) = 1

2. d/dx (sinx) = cosx

3. d/dx (cosx) = - sinx

4. d/dx (secx) = secx tanx

5. d/dx (any constant) = 0