Question

Find `dy/dx`if, y = `(x)^x + (a^x)`.

Answer 1

y = `(x)^x + (a^x)`

Let u = (x)^x and v = (a^x)

∴ y = u + v

Differentiating both sides w.r.t.x, we get

`dy/dx = (du)/dx + (dv)/dx`     ....(i)

Now u = `(x)^x`

Taking logarithm of both sides, we get

log u = log `(x)^x`

∴ log u = x . log x

Differentiating both sides w.r.t.x, we get

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

`= x * 1/x + log x * (1)`

∴ `1/u (du)/dx = 1 + log x`

∴ `(du)/dx = u(1 + log x)`

∴ `(du)/dx = (x)^x` (1 + log x)        ....(ii)

v = a^x

Differentiating both sides w.r.t.x, we get

`(dv)/dx = a^x* log a`       ....(iii)

Substituting (ii) and (iii) in (i), we get

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