Find `("d"y)/("d"x)`, if y = x^{(x)} + 20^{(x)}

**Solution:** Let y = x^{(x)} + 20^{(x)}

Let u = `x^square` and v = `square^x`

∴ y = u + v

Diff. w.r.to x, we get

`("d"y)/("d"x) = square/("d"x) + "dv"/square` .....(i)

Now, u = x^{x}

Taking log on both sides, we get

log u = x × log x

Diff. w.r.to x,

`1/"u"*"du"/("d"x) = x xx 1/square + log x xx square`

∴ `"du"/("d"x)` = u(1 + log x)

∴ `"du"/("d"x) = x^x (1 + square)` .....(ii)

Now, v = 20^{x}

Diff.w.r.to x, we get

`"dv"/("d"x") = 20^square*log(20)` .....(iii)

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

`("d"y)/("d"x)` = x^{x}(1 + log x) + 20^{x}.log(20)

#### Solution

Let y = x^{(x)} + 20^{(x)}

Let u = `x^x` and v = `20^x`

∴ y = u + v

Diff. w.r.to x, we get

`("d"y)/("d"x) = "du"/("d"x) + "dv"/("d"x)` .....(i)

Now, u = x^{x}

Taking log on both sides, we get

log u = x × log x

Diff. w.r.to x,

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

∴ `"du"/("d"x)` = u(1 + log x)

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

Now, v = 20^{x}

Diff.w.r.to x, we get

`"dv"/("d"x") = 20^x*log(20)` .....(iii)

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

`("d"y)/("d"x)` = x^{x}(1 + log x) + 20^{x}.log(20)