Question

Differentiate the function with respect to x:

x^x + x^a + a^x + a^a, for some fixed a > 0 and x > 0

Answer 1

Let, y = x^x + x^a + a^x + a^a

On differentiating with respect to x,

`dy/dx = d/dx (x^x) + d/dx (x^a) + d/dx (a^x) + (a^a) d/dx (1)`

= `d/dx (x^x) + ax^(a - 1) + a^x log a + 0`   ...(1)

Let, u = x^x 

Taking log on both sides,

log u = x log x

On differentiating with respect to x,

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

= `x * 1/x + log x`

= (1 + log x)

∴ `(du)/dx` = u (1 + log x)

= x^x (1 + log x)

i.e., `d/dx (x^x) = (du)/dx`

= x^x (1 + log x)

Putting the value of `d/dx (x^x)` in equation (1),

`dy/dx` = x^x (1 + log x) + ax^{a − 1} + a^x log a