Question

Differentiate the function with respect to x.

`(x cos x)^x + (x sin x)^(1/x)`

Answer 1

Let, y = `(x cos x)^x + (x sin x)^(1/x)`

Differentiating both sides with respect to x,

`dy/dx = (du)/dx + (dv)/dx`  ..(1)

Now, u = (x cos x)^x

Taking logarithm of both sides,

log u = log (x cos x)^x

log u = x log (x cos x)

Differentiating both sides with respect to x,

`1/u (du)/dx = x d/dx log (x cos x) + log (x cos x) d/dx (x)`

= `x * 1/(x cos x) d/dx (x cos x) + log (x cos x) xx 1`

= `1/(cos x) [x d/dx cos x + cos x d/dx (x)] + log (x cos x)`

= `1/(cos x) [x (- sin x) + cos x xx (1)] + log (x cos x)`   ...[∵ log_e mn = log_e m + log_e n]

= `- x (sin x)/(cos x) + (cos x)/(cos x) + log x + log cos x`

= −x tan x + 1 + log x + log cos x

`therefore (du)/dx` = u [1 − x tan x + log x + log cos x]

= (x cos x)^x [1 − x tan x + log x + log cos x]  ...(2)

Also, v = `(x sin x)^(1/x)`

Taking logarithm of both sides,

log v = `log (x sin x)^(1/x)`

log v = `1/x log (x sin x)`

Differentiating both sides with respect to x,

`1/v (dv)/dx = 1/x d/dx log (x sin x) + log (x sin x) d/dx 1/x`

= `1/x 1/(x sin x) * d/dx (x sin x) + log (x sin x) (-1) x^-2`

= `1/(x^2 sin x) [x d/dx sin x + sin x d/dx (x)] + (log x + log sin x)(-1) x^-2`

= `1/(x^2 sin x)` [x cos x + sin x] `- 1/x^2  log x - 1/x^2  log sin x`

= `1/x^2` [1 + x cot x − log (x sin x)]

`therefore (dv)/dx = v * 1/x^2` [1 + x cot x − log (x sin x)]

= `(x sin x)^(1/x) * 1/x^2` [1 + x cot x − log (x sin x)]    ...(3)

Putting the values ​​of from equation (2) and (3) in equation (1),

∴ `dy/dx = (du)/dx + (dv)/dx`

= `(x cos x)^x [1 − x tan x + log x + log cos x] + (x sin x)^(1/x) * 1/x^2 [1 + x cot x - log (x sin x)]`