Question

Find `"dy"/"dx"` if y = `"x"^"x" + ("7x" - 1)^"x"`

Answer 1

y = `"x"^"x" + ("7x" - 1)^"x"`

Let u = x^x  and v = `("7x" - 1)^"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) 

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

Also, v = (7x – 1)^x

Taking logarithm of both sides, we get

log v = log(7x - 1)^x

∴ log v = x. log(7x – 1)

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

`1/"v" * "dv"/"dx" = "x" * "d"/"dx" log ("7x" - 1) + log ("7x" - 1) * "d"/"dx"`(x)

`= "x" * 1/("7x" - 1) * "d"/"dx" (7"x" - 1) + log (7"x" - 1) * (1)`

∴ `1/"v" * "dv"/"dx" = "x"/(7"x" - 1) (7 - 0) + log (7"x" - 1)`

∴ `"dv"/"dx" = "v"["7x"/(7"x" - 1) + log(7"x" - 1)]`

∴`"dv"/"dx" = (7"x" - 1)^"x" ["7x"/(7"x" - 1) + log(7"x" - 1)]`      ....(iii)

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

`"dy"/"dx" = "x"^"x" (1 + log "x") + (7"x" - 1)^"x" [log(7"x" - 1) + "7x"/(7"x" - 1)]`