If y = sec(tan⁻¹ x), then find dy/dx at x = 1?

Given:

y = sec(tan^-1 x)

Find dy/dx at x = 1

Concept Used:

To differentiate y with respect to x, we use the chain rule.

The derivative of sec(u) is sec(u)tan(u), and the derivative of tan^-1(x) is 1 / (1 + x²).

Calculation:

Step 1: Start with the given equation: y = sec(tan^-1 x)

Step 2: Differentiate y with respect to x: dy/dx = d[sec(tan^-1 x)] / dx

⇒ dy/dx = sec(tan^-1 x)tan(tan^-1 x) × d[tan^-1(x)] / dx

Step 3: Substitute the derivative of tan^-1(x):

⇒ dy/dx = sec(tan^-1 x)tan(tan^-1 x) × (1 / (1 + x²))

Step 4: Simplify tan(tan^-1(x)):

tan(tan^-1 x) = x

⇒ dy/dx = sec(tan^-1 x) × x / (1 + x²)

Step 5: At x = 1:

tan^-1(1) = π/4 (since tan(π/4) = 1)

sec(tan^-1(1)) = sec(π/4) = √2

⇒ dy/dx = sec(π/4) × 1 / (1 + 1²)

⇒ dy/dx = √2 × 1 / 2

⇒ dy/dx = 1/√ 2

Conclusion:

∴ The value of dy/dx at x = 1 is 1/√ 2.