Question

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

Answer 1

y = x^x + a^x

Let u = x^x

Then log u = log x^x = x · log x

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

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

= `x * d/dx (log x) + (log x) * d/dx (x)`

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

∴ `(du)/dx = u (1 + log x)`

= x^x (1 + log x)    ...(1)

Now, y = u + a^x

∴ `dy/dx = (du)/dx + d/dx(a^x)`

= x^x (1 + log x) + a^x . log a