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 = 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)`     ...`[because 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)

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

Taking logarithm of both sides,

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

Putting the value of `(du)/dx` from equation (2) and `(dv)/dx` from (3) in equation (1),

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