Question

The value of `dy/dx` at x = `π/2`, where y is given by y = `x^sinx + sqrt(x)`, is ______.

Options

• `1 + 1/sqrt(2π)`

• 1

• `1/sqrt(2π)`

• `1 - 1/sqrt(2π)`

Answer 1

The value of `dy/dx` at x = `π/2`, where y is given by y = `x^sinx + sqrt(x)`, is `underlinebb(1 + 1/sqrt(2π)`.

Explanation:

Given, y = `x^sinx + sqrt(x)`

Let y₁ = x^sinx and y₂ = `sqrt(x)`

Now, y₁ = x^sinx

`\implies` log y₁ = sin x log x

Differentiating w.r.t. x, we get

`1/y_1 dy_1/dx = cos x log x + 1/x sin x`

`\implies dy_1/dx = x^sinx [cos x log x + 1/x sin x]`

`(dy_1/dx)_(x = π/2) = (π/2)^(sin  π/2) [cos  π/2 log  π/2 + 2/π sin  π/2]`

= `π/2 xx 2/π`

= 1

Now, y₂ = `sqrt(x)`

`\implies dy_2/dx = 1/(2sqrt(x))`

`(dy_2/dx)_(x = π/2) = 1/(2sqrt(π/2)) = 1/sqrt(2π)`

Since, y = y₁ + y₂

∴ At x = `π/2, dy/dx`

= `dy_1/dx + dy_2/dx`

`\implies dy/dx = 1 + 1/sqrt(2π)`