Question

Find `dy/dx` if y = (log x)^x + x⁵.

Answer 1

Given:

y = (log x)^x + x⁵

Let u = (log x)^x and v = x⁵

Then y = u + v

Differentiating w.r.t. x:

`dy/dx = (du)/(dx) + (dv)/(dx)`   ...(1)

To find `(du)/(dx)`:

u = (log x)^x

Taking log on both sides:

log u = log (log x)^x

log u = x · log (log x)

Differentiating w.r.t. x:

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

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

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

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

`(du)/(dx) = (log x)^x [1/(log x) + log(log x)]`   ...(2)

To find `(dv)/(dx)`:

v = x⁵

Differentiating w.r.t. x:

`(dv)/(dx) = 5x^4`   ...(3)

Substituting (2) and (3) in (1):

`dy/dx = (log x)^x [1/(log x) + log(log x)] + 5x^4`