Question

Derivative of cos⁻¹ `((sin x + cos x)/(sqrt2)), -pi/4 < x < pi/4` with respect to x is ______.

विकल्प

• −1

• 1

• `pi/4`

• `-pi/4`

Answer 1

Derivative of cos⁻¹ `((sin x + cos x)/(sqrt2)), -pi/4 < x < pi/4` with respect to x is −1.

Explanation:

Step 1: Simplify the expression

Split the fraction as follows:

`(sin x + cos x)/sqrt2 = 1/sqrt2 sin x + 1/sqrt2 cos x`

Since cos `(pi/4) = 1/sqrt2 and sin (pi/4) = 1/sqrt2`

Substituting these values:

sin `(pi/4) sin x + cos (pi/4) cos x`

Using the identity cos (A − B) = cos A cos B + sin A sin B:

`((sin x + cos x)/(sqrt2)) = cos x cos (pi/4) + sin x sin (pi/4)`

= cos `(x - pi/4)`

Step 2: Simplify the inverse function

Substitute this back into the original function:

y = `cos^-1 [cos (x - pi/4)]`

Given the interval `-pi/4 < x < pi/4`, we check the range of the argument:

`-pi/4 - pi/4 < x - pi/4 < pi/4 - pi/4`

`-pi/2 < x - pi/4 < 0`

Since cos(−θ) = cos(θ), we can write:

y = `cos^-1 [cos(pi/4 - x)]`

`0 < pi/4 - x < pi/2`, which falls within the principal value branch [0, π] of cos⁻¹.

`y = pi/4 - x`

Step 3: Differentiate with respect to x

`dy/dx = d/dx (pi/4 - x)`

`dy/dx = 0 - 1`

= −1

The derivative is −1.