Question

Differentiate the function with respect to x.

`x^(xcosx) + (x^2 + 1)/(x^2 -1)`

Answer 1

Let, y = `x^(x cos x) + (x^2 + 1)/(x^2 - 1)`

Again, let y = u + v

Differentiating both sides with respect to x,

`dy/dx = (du)/dx + (dv)/dx`   ...(1)

Now, u = `x^(x cos x)`

Taking logarithm of both sides,

log u = `log x^(x cos x)`

log u = x cos x log x

Differentiating both sides with respect to x,

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

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

= cos x + log x [x (−sin x) + cos x]

= cos x + x (−sin x) · log x + cos x · log x

`therefore (du)/dx = u [cos x log x - x sin x log x + cos x]`

= `x^(x cos x)` [cos x log x − x sin x log x + cos x]   ....(2)

Also, v = `(x^2 + 1)/(x^2 - 1)`

Differentiating both sides with respect to x,

`(dv)/dx = ((x^2 - 1) d/dx (x^2 + 1) - (x^2 + 1) d/dx(x^2 - 1))/((x^2 - 1)^2)`

= `((x^2 - 1)(2 x) - (x^2 + 1) (2 x))/((x^2 - 1)^2)`

= `(2 x [x^2 - 1 - x^2 - 1])/((x^2 - 1)^2)`

= `(-4x)/((x^2 - 1)^2)`  ....(3)

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

`therefore dy/dx = (du)/dx + (dv)/dx` 

`= x^(x cos x) [cos x log x - x sin x log x + cos x] - (4x)/(x^2 - 1)^2`