Question

Find `dy/dx` if y = x^x + 5^x

Answer 1

y = x^x + 5^x

let u= x^x and v = 5^x
∴ y = u + v

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

Now u = x^x
Taking logarithm on both the sides
log u = log x^x
⇒ log u = x.log x
    Differentiate w.r.t.x.
    `1/u (du)/dx = x. 1/x + log x.1`

                        = 1 + log x

`therefore dy/dx = u( 1 + log x )`

`therefore du/dx = x<sup>x</sup>( 1 + log x )`      (ii) 

Now, v = 5^x

`therefore dv/dx = 5<sup>x</sup>.log 5`               (iii)
Substituting (ii) and (iii) in eqn (i)

`dy/dx = x^x ( 1 + log x ) + 5^x log 5`